The velocity of water exiting nozzle is 14.2 m/s.
What is velocity ?
It measures how quickly a distance changes. It is the rate at which displacement is changing.
What is pressure?
Pressure is the force applied perpendicularly to an object's surface in relation to the area across which it is dispersed.
New same height so h₂=h₁
new position p₂= 101300 pa and velocity v₂
water density p₂= 1000kg/m
gravity = 9.8m/s
V₁²= 2/p (p₁-p₂)
4+2/1000 ( 20000- 101300 )
=20.14
v₂= [tex]\sqrt{20.14}[/tex]
= 14.2 m/s
Therefore, the velocity of water exiting nozzle is 14.2 m/s.
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in which of the following cases will there be a current in the ammeter? (1) the switch has just been closed (2) the switch has been closed a long time (3) the switch has just been opened (4) the switch has been opened a long time
The presence of a current in an ammeter depends on the state of an external electrical switch. If the switch is closed, there will be a current in the ammeter; if the switch is opened, there will be no current in the ammeter. The length of time the switch has been in a particular state does not affect the presence of a current in the ammeter.
The presence of a current in an ammeter depends on the state of an external electrical switch. A switch is a device used to break or make an electrical connection, and can be used to control the flow of electricity in a circuit. In the following cases, the presence of a current in the ammeter can be determined:
(1) If the switch has just been closed, there will be a current in the ammeter. When the switch is closed, it creates an electrical connection, allowing electricity to flow through the circuit, and thus a current to be registered on the ammeter.
(2) If the switch has been closed for a long time, there will still be a current in the ammeter. This is because the switch is still in the closed position, allowing electricity to flow through the circuit, and thus a current to be registered on the ammeter.
(3) If the switch has just been opened, there will be no current in the ammeter. When the switch is opened, it breaks the electrical connection, preventing electricity from flowing through the circuit, and thus no current will be registered on the ammeter.
(4) If the switch has been opened for a long time, there will still be no current in the ammeter. This is because the switch is still in the open position, preventing electricity from flowing through the circuit, and thus no current will be registered on the ammeter.
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A clean nickel surface (work function 5.1eV), is exposed to light of wavelength 206nm. What is the maximum speed of the photoelectrons emitted from this surface?
A pristine nickel surface with a work function of 5.1 eV is exposed to 206 nm light. The photoelectrons released from this surface have a maximum speed of 521.8 x 10^3 m/s.
An electron of the metal can receive photon energy from incident light, which is equal to the planck's constant times the light's frequency.
f = c/∧
assuming that all of the photon energy has been utilized
First in removing electrons from metals (work function)
Second, the leftover is used to furnish the photoelectron with kinetic energy:
V(m) = √2/M(hc/∧)-5.1eV.
where: M = 9.11 x10^3 kg; is the mass of nickel
h = 6.63 x 10 -34 Js; is plancks constant
c = 3 x 10^8 m/s; is the speed of light
∧ = wavelength of light = 206 nm = 206x10-9m
v(m) = (2 x 6.63 x 10-34 x 3 x10^8)/206 x 10-9 x 9.11x10^3 - 5.1 eV
v(m) = 521.8 x 10^3 m/s.
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60 m/s while piece c moves in the positive x-direction at 70 m/s. we want to find the speed and direction of piece b.
The speed and direction of piece b is 11.33m/s positive x direction.
Total momentum = 0 (at rest)
Applying conservation of linear momentum,
Momentum after explosion = momentum before explosion
MaVa + MbVb + McVc = 0
100(-60) + 150Vb + 70(86) = 0
Vb = 11.33m/s
Speed of Vb = 11.33m/s
In Newtonian mechanics, momentum is the result of an object's mass and velocity, more precisely linear momentum or translational momentum. This mechanical quantity is typical. It has. a direction. and a magnitude., making. it a vector. quantity. The momentum of an item is expressed as: [tex]displaystyle mathbf p =m mathbf v.mathbfp = m mathbf v[/tex] if m is the object's mass and v is its velocity (also a vector quantity).
The kilogram metre per second (kg m/s), which is a unit of measurement for momentum in the International System of Units (SI), is a unit of measurement for a newton-second.
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A 2.20-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. Part A What will be the speed of the upper end of the pole just before it hits the ground?
the speed of the upper end of the pole just before it hits the groundv is = 7.27 m / s
Explanation:
At both the greatest and lowest points, we shall practise energy conservation.
first. The highest point
Em₀ = U
Em₀ = m g y
final. smallest point
At this point, we adjust the reference system so that the potential energy is 0
= K
= ½ m v²
how to conserve energy
Em₀ =
mg y = ½ m v²
v = √2 gy
v = √ (2 9.8 2.7) (2 9.8 2.7)
v = 7.27 m / s
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a motorist drives south at 25m/s for 3.00 min, then turns west and travels at 20m/s for 2.60 min, and finally travels northwest at 28m/s for 1.00 min. For this 6.00min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive x axis point east.
a) D = -4273(x)-2327(y)
b) average speed v = 23.33m/s
c) average velocity V = -13.51m/s south of west.
In order to find the displacement we have to divide the motion on its X and Y components:
dx1 = 25*2*60min = 3000
dx2 = 30*1*60s/min(cos45) = 1273 m
The total X displacement is the algebraic sum of the movement:
dx = -3000 - 1273 = -4273
We have to do the same on Y:
dy1 = 20*3*60 = 3600
dy2 = 30*1*60(sin45) = 1273
dy = -3600 + 1273 =-2327m
D=-4273(x)-2327(y)
The angle is given by:
α = arctg(dy/dx)
α = 28.57 degrees
That is 28.57 degrees south of west.
the total displacement magnitude is given by:
D = √{(4273) - (2327)}
D= 4865
The average speed is given by:
v = (20*3)+(25*2)(1*60)/6*60
v =23.33 m/s
and the Average velocity is given by:
|v| = 4865/6*60
|v| =13.51 m/s
Therefore, the average velocity is V = -1351 m/s with an angle of 28.57 degrees south of west.
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a) D = -4273(x)-2327(y)
b) average speed v = 23.33m/s
c) average velocity V = -13.51m/s south of west.
In order to find the displacement we have to divide the motion on its X and Y components:
dx1 = 25*2*60min = 3000
dx2 = 30*1*60s/min(cos45) = 1273 m
The total X displacement is the algebraic sum of the movement:
dx = -3000 - 1273 = -4273
We have to do the same on Y:
dy1 = 20*3*60 = 3600
dy2 = 30*1*60(sin45) = 1273
dy = -3600 + 1273 =-2327m
D=-4273(x)-2327(y)
The angle is given by:
α = arctg(dy/dx)
α = 28.57 degrees
That is 28.57 degrees south of west.
the total displacement magnitude is given by:
D = √{(4273) - (2327)}
D= 4865
The average speed is given by:
v = (20*3)+(25*2)(1*60)/6*60
v =23.33 m/s
and the Average velocity is given by:
|v| = 4865/6*60
|v| =13.51 m/s
Therefore, the average velocity is V = -1351 m/s with an angle of 28.57 degrees south of west.
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two handheld radio transceivers with dipole antennas are separated by a large fixed distance. if the transmitting antenna is vertical, what percentage of the maximum received power will appear in the receiving antenna when it is inclined from the vertical by the following angles?
A transceiver is a combination transmitter/receiver in a single bundle. while the time period normally applies to wireless communications gadgets, it is able to also be used for transmitter/receiver devices in cable or optical fiber structures.
The electricity of the antenna is given by,
The resistance of the antenna is steady, for this reason,
Pa(avy
Now the voltage difference is,
AV = Bjcos8
consequently, the voltage distinction is directly proportional to t
cosine of the angle.
AVecosé (because B and/are constant)
subsequently, the electricity received,
Paces
therefore, if the angle of the willing is 14.three°, then the
acquired electricity by the antennas,
Pm c0s414,3°=0,93899
Paw
93.9%
if the perspective of the inclined is 45°, then the
obtained electricity by means of the antenna is,
SP price a5¢=05
Pan
50%
if the perspective of the inclined is 90°, then the
received energy with the aid of the antennas,
Pn cos? 9%=
Paw
= 0.0%
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Porosity decreases Choose one: A. with decreasing compaction of sediments or rock. B. when rocks develop joints or faults. C. when fluids passing through a rock dissolve parts of the rock. D. with the cementing of sediments by mineral grains from groundwater. The water table Choose one: O A. usually lies within a few meters of the surface in deserts. O B. rises to a higher elevation during the dry season. O C. mimics the topography of the land it underlies. O D. is always located beneath the ground surface.
A decrease in porosity occurs when rocks form joints or faults.
Option B is correct
What impact does porosity have on groundwater?Porosity is the entire available area for groundwater to live in. A rock's or sediment's porosity impacts how much water it can hold. The porosity of sediments and sedimentary rocks is influenced by the size, shape, and degree of sorting of the grains as well as the degree of cementation.
What impact do permeability and porosity have on groundwater?A good permeability must coexist with good porosity for groundwater to enter a rock. The pore gaps between the rock grains must be linked for a rock to be permeable and allow water to pass through it.
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multiple choice question the ratio that measures the reasonableness of accounts receivable outstanding, and can be used to estimate the average collection period of accounts receivable is the ratio.
The ratio that measures the reasonableness of accounts receivable outstanding, and can be used to estimate the average collection period of accounts receivable is the accounts receivable turnover ratio.
The amount owed to a business for goods or services that have been delivered or utilized, but for which consumers have not yet made payment is known as accounts receivable, or AR. A current asset is defined as accounts receivable on the balance sheet. Any sum that clients owe for purchases they made with credit is considered AR. Account receivable is the term used to describe any unpaid invoices or cash that a business is owed by customers. Accounts that a company is entitled to get as a result of delivering a good or service are referred to by the phrase. Receivables, also known as accounts receivable, are a type of line of credit that a business extends to customers. Typically, the terms of receivables stipulate that payments must be made within a reasonable amount of time.
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The drawing shows box 1 resting on a table, with box 2 resting on top of box 1. A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2, and the other end is connected to box 3. The weights of the three boxes are W1 = 51.5 N, W2 = 39.0 N, and W3 = 25.3 N. Determine the magnitude of the normal force that the table exerts on box 1.
The force that the table exerts on the box1 is 65.2N
As given in the question we are assuming box 2 and 3 aren't touching.
The forces on box 2 are:
gravitational force
Force of tension
Normal force
The box isn't accelerating so
T + N2 - Fg2 = 0
The forces acting on the box 3 are:
gravitational force
Tension
Since box 3 isn't accelerating
T - Fg3 = 0
The tension force will be same in both cases so:
Fg3 + N2 - Fg2 = 0
So
N2 = Fg2 -Fg3
N2 = Fg2 - Fg3
= 39-25.3 = 13.7 N
Remember forces are equal and opposite.
Now,
write the equation for the forces which are acting on Block 1
There is the force of gravity
normal force acting from the table and
A force which is equal and is opposite to the normal force on block 2
N1 - F(g1) - N2 = 0
N1 = Fg1 +N2
N1 = 51.5 + 13.7 = 65.2 N
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A member ABC has a mass of 2.4kg and is attached to pin support at B. An 800 gram sphere D strikes the end of the member ABC with a vertical downward velocity v1 = 3.0 m/s. knowing that L=750 mm and that the coefficient of restitution between the sphere and the Plank is 0.5. Determine immediately after impact:
a) The angular velocity of member ABC.
b)The velocity of the sphere.
(a)The angular velocity v(d)' is 0.936 m/s
(b)The velocity of the sphere is 3 rad/s
Let us assume that
angular velocity of member ABC is v(d)'
velocity of the sphere = ω'
Write the moment of inertia of the member ABC
I(G) = 1/12 ML²
I(G) = 1/12*(2.4)(0.75m)²
I(G) = 0.112 kgm²
After the impact
ω = ω'(ccw)
ν'(G) = (L/4)ω' (upwards)
ν'(A) = (L/4) ω' (downwards)
Take the moments about B
m(d)v(d)(L/4) + 0 = m(d)v'(d)(L/4) + I(G)ω' + Mv'(G)(L/4)
m(d)v(d)(L/4) = m(d)v'(d)(L/4) + I(G)ω' + M(L/4)²ω'
m(d)v(d)(L/4) = m(d)v'(d)(L/4) + ] I(G) + M(L/4)² ]ω'
(0.8)(3)(0.75/4) = (0.8)v'(d)(0.75/4) + [ 0.112 + (2.4)(0.75/4)²
0.45 = 0.15v(d)' + 0.1963ω' -------------------- (eq1)
From the coefficient of restitution,
v(d)' - v(A)' = -e [ v(d) - v(A) ]
v(d)' - (L/4)ω' = -(0.5) [3-0]
v(d)' - (0.75/4)ω' = - 1.5
v(d)' - 0.1875ω' = - 1.5 ---------------------- (eq 2)
From eq 1 and eq 2
v(d)' = 0.936 m/s
ω' = 3 rad/s
Therefore, the angular velocity is 0.936 m/s and the velocity of the sphere is 3 rad/s.
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15. A 5.0 g coin sliding to the right at 25.0 cm/s makes an elastic head-on collision with a 15.0 g coin that is initially at rest. After the collision, the 5.0 g coin moves to the left at 12.5 cm/s. Find the final velocity of the other coin. (12.5 cm/s, Right)
Answer: Finally, we can use this value to find the final velocity of the 15.0 g coin. Since velocity is defined as the change in position over time, and the 15.0 g coin is initially at rest, its final velocity must be equal to its final momentum divided by its mass. Therefore, the final velocity of the 15.0 g coin is 187.5 g cm/s / 15.0 g = 12.5 cm/s. Since the 5.0 g coin is moving to the right and the 15.0 g coin is moving
Explanation: To solve this problem, we can use the conservation of momentum. Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. In an elastic collision, momentum is conserved, meaning that the total momentum of the system before the collision is equal to the total momentum after the collision.
We can use this principle to find the final velocity of the 15.0 g coin. First, we need to calculate the initial momentum of the system. The 5.0 g coin has an initial momentum of 5.0 g * 25.0 cm/s = 125 g cm/s. Since the 15.0 g coin is initially at rest, its initial momentum is 0 g cm/s. The total initial momentum of the system is therefore 125 g cm/s.
After the collision, the 5.0 g coin moves to the left at 12.5 cm/s, so its final momentum is 5.0 g * (-12.5 cm/s) = -62.5 g cm/s. Since momentum is conserved, the final momentum of the system must be equal to the initial momentum of the system, so the final momentum of the 15.0 g coin must be 125 g cm/s - (-62.5 g cm/s) = 187.5 g cm/s.
Finally, we can use this value to find the final velocity of the 15.0 g coin. Since velocity is defined as the change in position over time, and the 15.0 g coin is initially at rest, its final velocity must be equal to its final momentum divided by its mass. Therefore, the final velocity of the 15.0 g coin is 187.5 g cm/s / 15.0 g = 12.5 cm/s. Since the 5.0 g coin is moving to the right and the 15.0 g coin is moving
The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. Suppose that a horizontal wind blows in streamline flow with a speed of 10.5 m/s outside a large pane of plate glass with dimensions 6.00 m Times 1.10 m. Assume the density of air to be constant at 1.20 kg/m^3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is 21 m/s, twice as high.
Given,
Speed of wind v = 10.5 m/s
Dimension of plate = 6 m times 1.10 m
The density of air d = 1.20 kg/m³
Therefore,
F = (1/2)dv²A
A = 6(1.10)m² [From dimension]
F = (1/2)(1.20)(10.5)²(6)(1.10)
F = 436.59 N
The force on the window pane is 436.59 N.
(a) if the wind speed is 21 m/s
F = (1/2)(1.20)(21)²(6)(1.10)
F = 1746.36 N
The force on the window pane is 1746.36 N.
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Consider the 52.0 kg mountain climber in the figure. 15° (a) Find the tension in the rope in N) and the force that the mountain climber must exert with her feet in N) on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. tension in rope force on feet ON (b) What is the minimum coefficient of friction between her shoes and the cliff?
a. the force that the mountain climber must exert with her feet in N) on the vertical rock face to remain stationary is 273 N.
b. the minimum coefficient of friction between her shoes and the cliff is 0.268.
What is the term of friction?
The force that prohibits one solid object from traveling across another is known as friction. Static friction, sliding friction, rolling friction, and fluid friction are the four main categories of friction.
a).
Balancing the forces in x-direction,
F cos(15) = T sin(31)
F = T sin(31)/cos(15) ...... (1)
balancing the forces in y-direction,
T cos(31) + F sin(15) = 52 x 9.8 = 509.6
T cos(31) + (T sin(31)/cos(15)) sin(15) = 509.6
T = (509.6 cos(15))/(cos(31) cos(15) + sin(31) sin(15))
T = 512 N
Force, F = 512 sin(31)/cos(15)
= 273 N
b)
Coefficient of friction,
u = sin(15)/cos(15)
= 0.268
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process command line args and get the simulation parameters (e.g., alg, quantum, inputfile) create/initialize the necessary data structures (ready q and io q (double linked lists of pcb),
In the operating system's command-line shell, parameters are provided after the program's name. Typically, we define main() with two arguments: a list of command-line arguments and a number representing the number of command-line arguments.
In C#, how do you acquire command-line arguments for a process?We'll talk about utilizing the Environment class to retrieve and show command-line arguments in this article. Therefore, we use the Environment Class's CommandLine property to do this task. To locate the command line for the active process, utilize this attribute. Message Type: This property returns a string as its return type.
Which of these methods accepts parameters from the command line?Only the main() method can accept parameters from the command line.
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Without air resistance, a projectile fired horizontally at 8 km/s from a mountain top will
a. return to its starting position and repeat its falling behavior.
b. accelerate downward at g as it moves horizontally.
c. trace a curve that matches Earth's curvature.
d. all of the above.
e. none of the above.d. all of the above.
Without an air resistance, a projectile fired horizontally at 8 km/s from a mountain top will:
Repeat the falling action by going back to your starting position. As it goes horizontally, accelerate downward at g. Draw a line that follows the curvature of the Earth.What is Gravity?All objects with mass or energy are drawn to one another by the basic interaction of gravity. Gravity is by far the weakest of the four fundamental interactions, and is greatly outweighed by the electromagnetic force, the weak interaction, and the strong interaction. As a result, it has little effect on phenomena occurring at the level of subatomic particles. However, gravity is the most significant interaction between objects at the macroscopic level and controls the motion of planets, stars, galaxies, and even light.
Similar to how gravity on the Earth gives weight to the physical objects, the Moon's gravity creates sublunar tides in the oceans (the corresponding antipodal tide is caused by the inertia of the Earth and Moon orbiting one another). Additionally, gravity is a key factor in variety of biological processes, such as gravitropism, which controls plant development, and fluid movement in multicellular organisms. Research on consequences of weightlessness indicates that the gravity may have an impact on how the immune system and cells differentiate in the human body.
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a permanent dipole and charged particle lie on the x-axis and are separated by a distance as indicated in the figure. the dipole consists of positive and negative charge separated by a distance where . the solo particle has a negative charge and is located at .
The vector quantity with a magnitude equal to the product of the two equal charges with opposing polarity separated by a distance, d, is defined as the electric dipole moment associated with it.
The distance between the charges and a direction running along the line between the charges from the negative to the positive charge.
Two charges totaling 10.0 C and spaced 2.00 cm apart make up the dipole. The positive charge is further away from the line of charge than the negative charge, and the dipole's axis forms a 35.0° angle with the x-axis.
The dipole will be subject to a force that is Charges 2.0108C separated by a distance of 2.0103m make up an electric dipole. As seen in the picture, it is positioned so that the negative charge is 2.0 cm away from a long line charge with a linear charge density of 4.0104Cm1. The pressure on the Axis.
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What device can be used to spot the differences between AC and DC?
A. Transformer
B. Generator
C. Motor
D. Oscilloscope
Answer:
Check explaination
Explanation:
A - Transformer is the right option 100%
evaporation is a cooling process and condensation isquestion 16 options:a cooling process also.a warming process.neither a warming nor cooling process.can not tell with the information given.
Evaporation is a cooling process and condensation is a warming process.
Condensation is the transition of a region of matter from a gaseous state to a liquid state and the opposite of vaporization. This word most often refers to the water cycle. Condensation is the process by which water vapor becomes a liquid. This is the opposite of evaporation, which turns liquid water into vapor. Condensation occurs in one of two ways. When the air cools to its dew point or becomes saturated with water vapor, it can no longer hold water.
Evaporation is always a cooling process. For evaporation to occur, energy is needed by the liquid molecules that are changing phase into gas. This energy is taken from the surroundings in the form of heat energy. So when evaporation occurs, the temperature of the surroundings will decrease, making evaporation a cooling process.
Condensation occurs when warmer, moist air comes into contact with cooler surfaces such as frames, windows, and other superstructures, or cooler areas inside an insulating shell (where moisture has penetrated the vapor barrier).
Therefore, evaporation is a cooling process and condensation is a warming process.
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a 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface.
The percentage of mechanical energy lost (converted into thermal energy) during the collision will be equal to 95.2%.
What is kinetic energy?Kinetic energy is the term used in physics to describe the force that a moving item has.
It is described as the amount of effort necessary to accelerate anybody with a particular mass from rest to a given velocity. Except for variations in speed, the body retains the kinetic energy it gains during acceleration.
As per the information in the question,
Ball's mass, m₁ = 50 g = 0.05 kg
Speed = v₀
Brick's Mass, m₂ = 1 kg
Assume the speed of the brick after the collision is v.
Use the equation of the principle of conservation of linear momentum,
[tex]m_1v_o+m_2u_2= v(m_1+m_2)[/tex]
0.5(v₀) + 0 = v(1 + 0.05)
v = (0.05v₀)/1.05
v = 0.048v₀.
Then the total kinetic energy will be:
K.E(initial) = 1/2 × 0.05v₀² + 1/2 × 0.05 × 0²
K.E(initial) = 0.025v₀²
Then,
K.E(final) = 1/2 × (1.05) × (0.048v₀)²
K.E(final) = 0.0012v₀²
So, the amount of mechanical energy lost in collision,
= (0.025v₀² - 0.0012v₀²)/0.025v₀² × 100
= 95.2%.
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This question seems incomplete, the complete question may be:
A 50 g ball of clay traveling at speed vo hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface. a. Find an expression for the speed of the brick after the collision in terms of vo. b. What percentage of the mechanical energy is lost (i.e. converted to thermal energy) in this collision?
Why is the following situation impossible? Figure shows Superman attempting to drink cold water through a straw of length â„“=12.0 m. The walls of the tabular straw are very strong and do not collapse. With his great strength, he achieves maximum possible suction and enjoys drinking the cold water.
(a) The maximum height to which Superman can lift the water is 10.32874 m
(b) On the Moon there is no atmosphere so no atmospheric pressure which means when the straw is placed in water water will not rise in the tube.
Pₐ = Atmospheric pressure = 101325 Pa
g = Acceleration due to gravity = 9.81 m/s²
h = Height of water
ρ = Density of water = 1000 kg/m³
If the walls of the tube do not collapse that means that maximum pressure inside will be the atmospheric pressure
Atmospheric pressure is given by
Pₐ = ρgh
height = Pₐ /ρg
h = 101325 / 1000 * 9.81
h = 10.328 m
The maximum height is 10.32874 m.
(b) On the Moon there is no atmosphere so no atmospheric pressure which means when the straw is placed in water water will not rise in the tube.
Therefore, the maximum height to which Superman can lift the water is 10.32874 m.
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[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: Superman attempts to drink water through a very long vertical straw. With his great strength, he achieves maximum possible suction. The walls of the tubular straw don't collapse.
(a) Find the maximum height through which he can lift the water.
(b) Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside the straw.]
rosa a block of mass 3 kg slides down an inclined plane at an angle of ??p with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline (see the following figure). the pulley can be approximated as a disk. the coefficient of kinetic friction on the plane is 0.4. what is the acceleration of the block? os 10
If the coefficient of kinetic friction on the plane is 0.4, the acceleration of the block is 0.3 m/s² and tension is T = 1.82N.
A property of a body that is a measure of inertia is usually viewed as a measure of the amount of matter contained in the body that causes it to have weight in a gravitational field. Mass is a dimensionless quantity that describes the amount of matter in a particle or object.
Calculation:-
Tension force = I/R² × a
= 0.5 × 3.564
T = 1.782 N.
acceleration = 1.782/3
= 0.3 m/s²
The mass of an object is a measure of its inertial properties or the amount of matter it contains. The weight of an object is a measure of the force exerted on it by gravity, or the force required to support it. Mass can best be understood as the amount of matter present in an object or body. Everything around us has mass.
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A ship sends sonar and receives an echo after 2 seconds. If the speed of sound waves in water is 1500 m/s, then the depth of the sea is
A 1500 m
B 500 m
C 1000 m
D 750 m
After 2seconds of sending sonar, a ship hears an echo. The depth of the sea is 1500 meters if the sound wave speed in water is 1500 m/s.
Let sea depth equal d.
Considering that the sound wave's water speed (v) is 1500 m/s
The sound must travel to the ocean's bottom before being reflected back to the ship, resulting in a total distance of two dimensions for the sound.
Time (t) = 2s for receiving the echo
Total distance traveled times the speed of sound
1500 m/s is equal to 2d/2s = 1500 m.
Consequently, the water is 1500 meters deep.
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An engineer wanted to put a speed limit at the enterance to atrain at 500maway as it comes to rest to a plant form his plan is give exactly 3/2 minutes for the plate to decelerate uniformaly within the 500m distance what must bethe maximum speed at the entrance?
Answer:
50
Explanation:
100×50÷1000= 50
this is easy question
Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Reset Help UR Vc + UL At any instant; vis equal to 909 But vC and UL are out of phase SO UC can be larger than v at & value of t, positive negative at that t vL + UR UR - "C UL UR + Uc + UL UL ~ UR 1800 UR + vc - VL
The voltages across the capacitor and inductor may be greater than the input voltage when the circuit is at or close to the resonance. The resistance voltage is 90 degrees behind the inductor voltage.
How do you determine a capacitor's voltage and amplitude?The VC (t) phasor has an amplitude of VC=I0XC and lags the i(t) phasor by /2 rad. The vL(t) phasor has the amplitude VL=I0XL and trails the i(t) phasor by /2 rad.
Is it possible for the voltage across the capacitor to be higher than the voltage at the source?In a series LCR circuit, the voltage drop across the inductor or capacitor can, in fact, exceed the applied voltage of the a.c. source. VC or VL may be bigger than V since VC and VL have opposing faces. The RMS current for an applied V = volts RMS voltage is I = x 10 amperes. You may calculate the component voltages by multiplying the current.
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If 0 > 0 means anti-clockwise rotation, then an object with angular velocity w < 0 and angular acceleration a < 0 means the object is O rotating in the anti-clockwise direction and slowing down O rotating in the clockwise direction and speeding up O rotating in the clockwise direction and slowing down The centripetal (normal) acceleration is related to O the angular velocity as angular velocity times radius O none of the rotational (angular) parameters O the angular acceleration A top is spinning counterclockwise on a flat table. Its angular momentum vector is O An arrow pointing up Zero An arrow parallel to the surface of the table An arrow pointing down
Its angular momentum vector of it is O.A. downward-pointing arrow with a clockwise motion and accelerating speed. no rotational (angular) characteristics. an upwardly directed arrow.
1. A negative angular velocity indicates that motion is in the clockwise direction. The object is speeding up because the signs of angular acceleration and velocity are both negative. Thus, turning in a clockwise direction and quickening.
2. The standard acceleration is equal to v2/R. Consequently, none of the rotational (angular parameters)
3. An arrow pointing up represents the top's angular momentum as it spins counterclockwise.
An object's or system of objects' angular momentum is a property that describes its rotational inertia as they move around an axis that may or may not pass through them. Due to its yearly rotation around the Sun and daily rotation about its axis, the Earth has both orbital and spin angular momentum. An accurate explanation of angular momentum necessitates the identification of both a magnitude and a direction because it is a vector quantity. An orbiting object's angular momentum is proportional to its linear momentum, which is the sum of its mass m and linear velocity v, multiplied by the perpendicular distance r from the center of rotation to a line drawn across its instantaneous motion.
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The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. Suppose that a horizontal wind blows in streamline flow with a speed of 10.5 m/s outside a large pane of plate glass with dimensions 6.00 m Times 1.10 m. Assume the density of air to be constant at 1.20 kg/m^3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is 21 m/s, twice as high.
Force is 436.59 N and the speed is 1.746 kN.
A) Force F = P*A = 0.5*rho*v^2*A = 0.5*1.2*10.5^2*6*1.1 = 436.59 N
B) If v is doubled then F increases by 4 times since F is proportional to the square of the speed.
then F2 = 4*436.59 = 1746.35 N = 1.746 kN
A body can change its condition of rest or motion through the application of force, which is an external agent. It has. both a magnitude. and. a direction. The application of force is the point at which force is applied, and the direction in which the force is applied is referred to as the direction of the force. If motion is conceivable, gravitational forces can separate phases of various densities. One option is widespread melting, but high temperatures are not anticipated for the early Earth (Chapter 2). Sorting as a result of deeply buried ices outgassing is a more likely explanation. Such processes would have altered the moment of inertia, necessitating a modification of angular velocity.
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Five different wind shear profiles are shown below. Please place a triangle around the letter beneath the wind shear profile that is ideal for a supercell. Please place a circle around the letter beneath the wind shear profile that is ideal for a squall line 200 mb 250 mb 300 mb 400 mb 500 mb 600 mb 700 mb 850 mb 000 mb A. C. Wind shear is calculated by finding the difference between the wind speed values at two different heights. Using the wind profile you identified as the ideal environment for a supercell, next you will calculate the 0-1 km and 0-5 km wind shear values. This means we will find the difference between the surface wind (lowest wind barb on the sounding) and the speed of the wind at lkm and 5 km. The atmospheric pressure at 1km above sea level is typically very close to 850 mb. The pressure at 5 km above sea level is very close to 500 mb. Please calculate the 0-1 km and 0-5 km wind shear values in knots (kts). 0-1km shear value 0-5km shear value
At the surface, the wind speed is between 13 to 17 knots. At 1 km or 850 MB level, the wind speed is between 48 to 52 knots. Thus, the 0-1 km wind shear is (48 -13) or (52-17) = 35 knots.
At 5 km or 500 MB level, the wind speed is between 68 to 72 knots. Thus, the 0-5 km wind shear is (68-13) or (72-17) = 55 knots.
Wind speed describes how fast air moves past a particular point. This can be averaged over specific time units. B. Miles per hour or instantaneous speed is reported as peak wind speed gust or squall. The wind is the natural movement of air or other gases against the surface of a planet.
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BONUS: Make a sketch and label all the forces. A 2.00 kg box slides down an inclined plane
(ramp) at an angle of 20.0 degrees at a constant velocity. If the horizontal length of the
inclined plane is 2.00 m (you need to determine the actual slope length or hypotenuse):
(a) Calculate the work done by the force of gravity.
(b) Calculate the work done by the force of kinetic friction.
By signing my name to the test below, I pledge that the work that I have turned in for this
test is entirely my own and I have neither given nor received help from anyone else.
The amount of labour that gravity does is 28.58J. And the work produced by kinetic friction is -28.58J. A force that acts between moving surfaces is referred to as kinetic friction.
We may determine the slope length or hypotenuse using elementary trigonometry: cos(20) = 2.00. (slope length)
slope length: 2.00 m/cos(20) = 2.13 m
The box covers this distance while moving at a constant speed.
In terms of the forces, the total force parallel to the inclined plane acting on the box is given by the equation: F = W - f = mgsin(20) - f, where f represents friction and W is weight.
Work = Wd = (+13.2N)(2.13m) = 28.58J, which is the amount of work done by the force of kinetic friction.
The force acts in the same direction as the displacement, resulting in work.
The kinetic friction force exerts the following work:
Work= -fd =(-13.42N)(2.13m) = -28.58J
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A car driving at 20 m/s accelerates 2.5 m/s2 for a distance of 350 meters. What is the car's final velocity?
The time of the travel with 20m/s is 17.5 seconds. Thus, the final velocity of the car will be 63 m/s.
What is velocity?Velocity of an object is the measure of the distance travelled per unit time. The rate of change in velocity is called the acceleration.
Given that, initial velocity u = 20 m/s
distance = 350 meters.
time of travel = distance/ velocity
= 350 m/20 m/s = 17.5 s.
Acceleration = 2.5 m/s²
the final velocity v = u + at.
v = 20 + (2.5 × 17.5 ) = 63 m/s.
Therefore, the final velocity of the car is 63 m/s.
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a student is asked to determine the mass of jupiter. knowing which of the following about jupiter and one of its moons will allow the determination to be made? i. the time it takes for jupiter to orbit the sun ii. the time it takes for the moon to orbit jupiter iii. the average distance between the moon and jupiter
By knowing i. the time it takes for Jupiter to orbit the sun and ii. the time it takes for the moon to orbit Jupiter's mass of Jupiter can be determined. Therefore options I and II are correct.
To determine the mass of Jupiter, the student needs information about both Jupiter's orbit around the Sun (I) and the moon's orbit around Jupiter (II).
By knowing the time it takes Jupiter to orbit the Sun (I) and the time it takes for the moon to orbit Jupiter (II), the student can use Kepler's Third Law of Planetary Motion.
This law relates the orbital period and the average distance of a moon from its planet to the mass of the planet.
By knowing both the orbital periods of Jupiter and its moon (I and II) and the average distance between the moon and Jupiter (III is not required), the student can calculate the mass of Jupiter using Kepler's Third Law.
Therefore options I and II are correct.
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