Answer:
0.83 m³/s
Explanation:
The speed of the airflow is given as;
1.85 x 10³ ft³/min
Now we are to express this unit in m³/s
1ft = 0.3m
60s = 1 min
So;
1.85 x 10³ x ft³ x [tex]\frac{1}{min}[/tex] x [tex]\frac{(0.3m)^{3} }{ft^{3} }[/tex] x [tex]\frac{1min}{60s}[/tex]
= 0.83 m³/s
Explain how momentum is determined and conserved.
ASAP!!
Answer: momentum is determined and conserved.
Answer:
monument is determined
At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s2
Answer:
(a) The two balls collide [tex]2\; \rm s[/tex] after launch.
(b) The height of the collision is [tex]4\; \rm m[/tex].
(Assuming that air resistance is negligible.)
Explanation:
Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.
The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].
Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].
On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].
Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:
[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].
For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].
For the ball thrown downwards:
Initial velocity: [tex]v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 40\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Similarly, for the ball thrown upwards:
Initial velocity: [tex]v_0 = \left(12\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 0\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Equate the two expressions and solve for [tex]t[/tex]:
[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].
[tex]t = 2[/tex].
Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.
Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:
[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].
In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].
The time the two balls collide is 0.4 seconds while the height at which they collide is 4m
The given parameters are :
Initial Velocity U = 8m/s
Height H = 40m
For the second ball, the initial velocity = 12m/s
a.) For the first ball, the height attained at the point of collision will be
h = ut + 1/2gt^2
h = 8t + 1/2 x 10t^2 ........ (1)
For the second ball, the height attained at the point of collision will be
h = 12t - 1/2 x 10t^2 .........(2)
Since the height will be the same for the two balls, equate the two equations
8t + 10t^2 = 12t - 10t^2
Collect the like term
8t - 12t = -5t^2 - 5t^2
-4t = -10^2
10t = 4
t = 4/10
t = 0.4s
b.) Substitute time t in any of the equation to find the height
h = 12(0.4) - 0.5 x 10(0.4)^2
h = 4.8 - 0.8
h = 4m
Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m
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Consider a simple pendulum consisting of a massive bob suspended from a fixed point by a string. Let T denote the time (the period of the pendulum) that it takes the bob to complete one cycle of oscillation (the time it takes for the pendulum to swing back and forth one time). How does the period of the swing of the simple pendulum depend on the quantities that define the pendulum and the quantities that determine the motion
Answer:
The period of the swing depends on only the length of the string and not on the mass of the bob and the period of the pendulum depends on only the horizontal component of g.
Explanation:
The period of the swing depends on only the length of the string and not on the mass of the bob. Since the length of the string and the mass of the bob define the pendulum.
Also, the properties that define the motion are the component of the weight of the bob in the horizontal direction which determines the to and fro movement of the bob. So, the period of the pendulum depends on only the horizontal component of g.
So, T = 2π√(l/g) where l = length of pendulum and g = acceleration due to gravity.
Find the value of F1 + F2 + F3.
Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
For F₁
[tex]F_{y}=2[N][/tex]
For F₂
[tex]F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N][/tex]
For F₃
[tex]F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N][/tex]
Now we can sum each one of the forces in the given axes:
[tex]F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N][/tex]
Now using the Pythagorean theorem we can find the total force.
[tex]F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N][/tex]
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
what is the formula of moment of force
I need help with number 3!!!!!!!!!!!!
Answer:
3. The frequency of the wave is 3 Hz.
Explanation:
3. Determination of the frequency of the wave.
Frequency is simply defined as the number of complete circle or oscillation made in 1 seconds. Mathematically, it can be expressed as:
f = n / t
Where:
f => is the frequency.
n => is the number of circle.
t => is the time.
With the above formula, we can obtain the frequency of the wave as follow:
Number of complete circle (n) = 3
Time (t) = 1 s
Frequency (f) =?
f = n / t
f = 3 / 1
f = 3 /s = 3 Hz
Therefore, the frequency of the wave is 3 Hz
In the diagram at the right, each grid square is 1.0μmx 1.0 μm. If Mass A = Mass C = 10kg and Mass B = 28 kg, determine the gravitational field at the location marked by the red dot.
Answer:
B
Explanation:
8-2 is 6
times that times 8 u get B
David is driving a steady 28.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.90 m/s^2 at the instant when David passes.
Required:
a. How far does Tina drive before passing David?
b. What is her speed as she passes him?
Answer:
Explanation:
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1
For David u = 28.0 m/s where 'a' is set to nought
S = ut
S = 28t.......2
For Tina consider equation 1
Where acceleration = 2.90m/s^2 and u is set at nought
S = 1/2×2.90 m/s×t^2.......3
Equate 2 and 3
28t = 1.45t^2
Divide through by t
28 = 1.45t
t = 28/1.45
t = 19.31seconds
Now put the value of t into equation 3
S = 1/2×2.90 m/s×t^2.......3
= 1.45×20×20
= 580m
Tina must have driven 580meters before passing David
Considering the equation of linear motion : V^2 = U^2+2as
Where u is set at nought
V^2 = 2as
V^2 = 2×2.9×580
V^2 = 3364
V = √3364
V = 58m/s
Her speed will be 58m/s
(a) Tina should drive for 580 m, before passing the David.
(b) The speed of Tina during her passage through the David is 58 m/s.
Given data:
The initial velocity of the David is, u = 28.0 m/s.
The magnitude of acceleration is, [tex]a = 2.90 \;\rm m/s^{2}[/tex].
(a)
We can use the second kinematic equations of motion to obtain the distance covered by Tina, before passing the David. As per the second kinematic equation of motion,
[tex]s= u't + \dfrac{1}{2}at^{2}[/tex]
Here, u' is the initial speed of Tina and t is the time interval. Then,
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion as,
S = ut + 1/2at²...............................................................(1)
Also,
S = ut
S = 28t ...........................................................................(2)
For Tina consider equation 1
S = 1/2×2.90t²................................................................(3)
Equate 2 and 3
28t = 1.45t²
28 = 1.45t
t = 28/1.45
t = 19.31 seconds
Now put the value of t into equation (3)
S = 1/2×2.90 t².
= 1.45×20×20
= 580m
Thus, we can conclude that Tina should drive for 580 m, before passing the David.
(b)
Now, using the third kinematic equation of motion to obtain the speed of Tina during her passage through David as,
v² = u²+2as
Solving as,
v² = 28.0² + 2(2.90)(580)
v = √3364
v = 58m/s
Thus, we can conclude that the speed of Tina during her passage through the David is 58 m/s.
Learn more about the Kinematic equation of motion here:
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Which factor/s affect the amount of energy stored in a magnetic field?
The answer is "I, II, and III only"
Answer:
The answer is I
Explanation:
Answer:
4.) l, ll, and lll only
Explanation:
The rest of the answers:
1.) The field energy will increase.
2.) The energy increases, and the lines of force are denser.
3.) It points toward the field of earths magnetic poles.
5.) ll, lV, l, lll
Two 13.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 V battery. What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process. What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process. What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process.
Answer:
Explanation:
The capacitor is of parallel plate capacitor type
Capacitance C = ε₀ A / 4π d
ε₀ is 8.85 x 10⁻¹² , A is plate area and d is distance between plate .
d = 1.7 cm
C = 8.85 x 10⁻¹² x π x (6.5 x 10⁻² )² / 4π x 1.7 x 10⁻²
= 55 x 10⁻¹⁴F .
Charge on each electrode = C x V , V is voltage of battery .
= 55 x 10⁻¹⁴ x 14
= 770 x 10⁻¹⁴ C
Electric field strength = V / d where V is potential difference of battery , d is distance between plate .
= 14 / 1.7 x 10⁻²
= 8.23 x 10² V / m
The potential difference between plate
= potential difference of the battery
= 14 V .
An air-filled parallel-plate capacitor is charged and then disconnected from the battery. The plates are then pulled apart to twice their original separation. Which of the following statements about this capacitor is true?
A. The capacitance has doubled.
B. The energy stored in the capacitor has doubled.
C. The potential difference across the plates has decreased.
D. The electric field between the plates has increased.
Answer:
B.
Explanation:
From the given information, since the capacitors are disconnected from the battery, the electric field between the plate does not change due to the fact that there is no difference in charge density.
Hence, the potential difference after separating the plates is:
V' = Ed
V = e (2d')
V' = 2V
For the energy stored in the capacitor;
[tex]U' = \dfrac{1}{2}QV'[/tex]
where;
V' = 2V
[tex]U' = \dfrac{1}{2} Q (2V)}[/tex]
[tex]U' =2( \dfrac{1}{2} Q V)[/tex]
U' = 2U
Thus, in the capacitor, the energy that is being stored is doubled.
10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends soccer ball towards the goal.
G
U
E
S
S
Formula
11) Small rockets are fired to make small adjustments in the speed of a satellite. A certain small rocket can change the velocity of a 72,000kg satellite from 0m/s to 0.63m/s in 1296s. What force is exerted by the rocket on the satellite?
G
U
E
S
S
Formula
please I need help I don't understand it and I had to deliver it yesterday helpp:(
Answer:
10. 65 N
11. 35 N
Explanation:
10. Determination the force that sends the soccer ball towards the goal.
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Initial velocity (u) of the ball = 0 m/s
Final velocity (v) of the ball = 32.5m/s time (t) = 0.21 s
Acceleration (a) of the ball =?
a = (v – u) /t
a = (32.5 – 0) / 0.21
a = 32.5 / 0.21
a = 154.76 m/s²
Finally, we shall determine the force that sends the soccer ball towards the goal. This can be obtained as follow:
Mass (m) of the ball = 0.42 kg
Acceleration (a) of the ball = 154.76 m/s²
Force (F) =?
F = ma
F = 0.42 × 154.76
F = 65 N
Thus, the force that sends soccer ball towards the goal is 65 N
11. Determination of the force exerted by the rocket on the satellite.
We'll begin by calculating the acceleration of the satellite. This can be obtained as follow:
Initial velocity (u) of satellite = 0 m/s
Final velocity (v) of satellite = 0.63 m/s
Time (t) = 1296 s
Acceleration (a) of the satellite =?
a = (v – u) /t
a = (0.63 – 0) / 1296
a = 0.63 / 1296
a = 4.861×10¯⁴ m/s²
Finally, we shall determine the force exerted by the rocket on the satellite. This can be obtained as follow:
Mass (m) of the satellite = 72000 Kg
Acceleration (a) of the satellite = 4.861×10¯⁴ m/s²
Force (F) =?
F = ma
F = 72000 × 4.861×10¯⁴
F = 35 N
Thus, the force exerted by the rocket on the satellite is 35 N
*PLEASE HELP*
When an object is placed in front
of a convex lens, it creates a virtual
image at -12.8 cm with a
magnification of 2.85. What is the
focal length of the lens?
(Mind your minus signs.)
(Unit = cm)
Answer: The focal length of the lens = 3.32 cm.
Explanation:
Lens formula : [tex]\dfrac1f=\dfrac1v-\dfrac1u[/tex] (i)
f= focal length , v=image distance , u =object distance.
magnification: m = [tex]\dfrac{v}{u}[/tex] (ii )
Given: v= -12.8 cm , m =2.85
Put values in (ii), we get
[tex]2.85=\dfrac{-12.8}{u}\\\\\Rightarrow\ u=\dfrac{-12.8}{2.85}\\\\\Rightarrow\ u=-4.49\ cm[/tex]
substitute values of u , v in (i)
[tex]\dfrac1f=\dfrac1{-12.8}-\dfrac{1}{4.49}\\\\\Rightarrow\dfrac1f=-0.30084214922\\\\\Rightarrow\ f=\dfrac{-1}{-0.30084214922}\approx3.32\ cm[/tex]
Hence, the focal length of the lens = 3.32 cm.
Answer:
6.92
Explanation:
2.85=-(-12.8/x)
do=4.49
1/f= 1/4.49 + 1/-12.8
f=6.92
The force of gravity acting on an object is directed through this
center of gravity and toward the center of the
Explanation:
Every object has a center of gravity. ... The force of gravity acting on an object is directed through this center of gravity and toward the center of the earth. The object's weight, W, can be represented by a vector directed down (along the line the object would fall if it were dropped).
It should be towards the center of the earth.
The following information should be considered:
Each and every object contains the center of gravity. The force of activity acted on an object that director via his gravity center & towards the center of the earth.Learn more: brainly.com/question/17429689
A ray of monochromatic light is incident on a plane mirror at and angle of 30. The angle of reflection for the light is
1)15
2)30
3)60
4)90
Answer:
30 degrees
Explanation: reflection, same angle
For a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
Reflection occurs when radiation bounces off from a surface. Light is an electromagnetic wave and it can be reflected. According to the laws of reflection, the angle of incidence is equal to the law of reflection.
Hence, for a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
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A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s 2 for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50 m/s 2 until it stops at the next station. Find the total distance covered.
Answer:
1796.48 m
Explanation:
Given that :
First part of journey :
Initial Velocity, u = 0
Acceleration, a, = 1.60 m/s²
Time, t = 14 s
Distance traveled, S = 0.5at²
S = 0.5 * 1.60 * 14²
S1 = 156.8m
2nd part :
Speed is constant
Time = 70 seconds
At constant speed ;
Distance = speed * time
Speed, V = u + at
V = 0 + 1.6*14
V = 22.4 m/s
Distance, S2 = 22.4 * 70 = 1568 m
3rd part :
Deceleration = - 3.50m/s²
Final velocity, v = 0
Time taken to attain rest
V = u + at
0 = 22.4 - 3.5(t)
3.5t = 22.4
t = 22.4/3.5
t = 6.4 seconds
S3 = ut - 0.5at² (deceleration)
S3 = (22.4*6.4) - 0.5(3.5)*6.4^2
S3 = 71.68m
S1 + S2 + S3
156.8m + 1568m + 71.68m
= 1796.48 m
Claim:
Elements are pure substances made up of one type of atom,
and are written as their chemical
are pure substances made up of bonded elements, and are written as a
Explanation:
Elements are distinct substances that cannot be split-up into simpler substances. These substances consist of only one kind of atom.
There are over one hundred elements known to date. Each of these elements is usually symbolized by a capital letter or a capital letter followed by a small letter derived from English or Latin or Greek name of the element concerned.
Elements can be categorized in different ways.
Atoms are the smallest unit of elements that takes part in a chemical reaction.
Two identical 0.25 kg balls are involved in a head-on collision. Ball A is initially travelling at 3.5 m/s, and ball B is initally at rest. Determine the velocity of each ball after the collision.
Answer:
a) mv(final):<0,0,0> minus mv(initial):<25,0,0> = <-25,0,0>
b) mv(final):<25,0,0> minus mv(initial):<0,0,0> = <25,0,0>
c) conservation of momentum makes it <0,0,0>
for a-b-c, momentum_system + momentum_surroundings = 0
Explanation:
Hope this helps
The velocity of each ball after the collision is 1.75 m/s
Law of conservation of momentum states that:
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Where m₁, m₂ is the mass of object, u₁, u₂ is the initial velocity before collision and v is the final velocity after collision
Given that: m₁ = m₂ = 0.25 kg, u₁ = 3.5 m/s, u₂ = 0, hence:
0.25(3.5) + 0.25(0) = (0.25 + 0.25)v
v = 1.75 m/s
The velocity of each ball after the collision is 1.75 m/s
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5.
What is the apparent colour of a red shirt when viewed in pure green light.?
Red
(b)- Green
Yellow (d) Black) (e) Blue
Answer: black
Explanation: When green light is shone on a red object, it absorbs all of the green light and not reflecting anything. Hence, it appears black.
An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended into heat for warming water. How much mechanical energy is required to increase the temperature of 300 g of water (enough for 1 cup of coffee) from 20°C to 95°C? (1 cal = 4.186 J, the specific heat of water is 4186 J/kg⋅°C)
Answer:
Emec = 94050 [J]
Explanation:
In order to solve this problem, we must understand that all thermal energy is converted into mechanical energy.
The thermal energy can be calculated by means of the following expression.
[tex]Q=m*C_{p}*(T_{final}-T_{initial})[/tex]
where:
Q = heat [J]
Cp = specific heat of water = 4186 [J/kg*°C]
m = mass = 300 [g] = 0.3 [kg]
T_final = 95 [°C]
T_initial = 20 [°C]
Now we can calculate the heat, replacing the given values:
[tex]Q=0.3*4180*(95-20)\\Q= 94050[J][/tex]
Since all this energy must come from the mechanical energy delivered by the exercise bike, and no energy is lost during the process, the mechanical energy must be equal to the thermal energy.
[tex]Q=E_{mec}\\E_{mec}=94050[J][/tex]
A system experiences a change in internal energy of 36 kJ in a process that involves a transfer of 14 kJ of heat into the surroundings. Simultaneously, which of the following is true?
a. Q= 14 u=36
b. U=Q-W
c. 14-36=22
d. 22 kJ of work is done on the system.
Answer:
b. U = Q - W
Explanation:
Given;
change in internal energy, ΔU = 36 kJ
heat transferred to the surroundings Q = 14 kJ
Apply first law of thermodynamic; the change in internal energy is equal to heat added to the system minus work done by the system.
ΔU = Q - W
Since heat was lost to surroundings, Q = - Q
ΔU = (-Q) - W
36 kJ = -14 kJ - W
36 kJ + 14 kJ = - W
50 kJ = - W
W = - 50 kJ (the negative sign shows that work has been done on the system)
Thus, 50 kJ of work is done on the system.
The only correct answer in the given options is "b" U = Q-W
a group of students working in a high school chemistry lab believe they have discovered a new element! how exciting! upon further testing by scientists (with better equipment),it is found that the element contains 74 protons and 110 neutrons.
Explanation:
From the experiment:
Number of protons = 74
Number of neutrons = 110
Number of protons in an element is the atomic number of the element. It is used to locate and position and element on the periodic table.
For a neutral or uncharged atom, the number of protons is the same as the number of electrons.
The element whose number of protons or atomic number if 74 is Tungsten
Mass number = 74 + 110 = 184g/mol
A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
what happens if we add a lump of ice to a tumbler containing water? how does heat flow?
Answer:
The ice will melt
Heat flows from the water to the ice
Explanation:
If we add a lump of ice to a tumbler containing water, the ice lump will melt in the water.
This is a simple physical change of state from solid to liquid.
At the end both the solid ice and the original liquid maintains the liquid state.
The heat will flow from the ice to the water in the tumbler. Heat generally flows from a body at higher temperature to one with a lower temperature. The water has a higher temperature when compared to the ice. So, heat flows from the liquid water to the ice until thermal equilibrium is reached.Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 to reach a building 41.0m away. You can ignore air resistance; assume that the end of the hose is at ground level.
Required:
a. Find the angle of elevation of the hose.
b. Find the speed in m/s of the water at the highest point in its trajectory.
c. Find the acceleration in m/s^2 of the water at the highest point in its trajectory.
d. How high above the ground in m does the water strike the building?
e. How fast is it moving in m/s just before it hits the building?
Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:vₓ₀ = v * cos θ (1)where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:[tex]x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)[/tex]
Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:[tex]cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)[/tex]
⇒θ = cos⁻¹ (0.526) = 58.3º (4)b)
At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)c)
At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)d)
Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)[/tex]
Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)Replacing (7) in (6), we get:[tex]\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)[/tex]
e)
When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.The horizontal component, since it keeps constant, is just v₀x:v₀ₓ = 13.7 m/sThe vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:[tex]v_{fy} = v_{oy} - g*t (9)[/tex]
Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:[tex]v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)[/tex]
Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:[tex]v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)[/tex]
a 1 mole of an ideal gas is kept at 0°C during expansion from 30l to 10l .How much work is done on the gas during expansion
Answer:
20 J
Explanation:
Work done is given force by distance .
W= F * d where F is force given by the product of pressure and area
W= P* Δv where Δv is change in volume.
Given that ;
1 mole of an ideal gas is kept at 0°C, the pressure of the gas is : 1 atm.
Δv is change in volume , 30 l - 10l = 20 l
W= 1 * 20 = 20 J
I’m confused how to start it and I just need help atleast doing one.
Answer:
CuCl₂ + H₂S → CuS + 2HCl
Explanation:
The unbalanced reaction expression is given as;
CuCl₂ + H₂S → CuS + HCl
The problem here involves balancing of chemical equations.
We use a mathematical approach to solve this problem. Here assign coefficients a, b, c and d as values that will effect the balance;
aCuCl₂ + bH₂S → cCuS + dHCl
Conserving Cu: a = c
Cl: 2a = d
H: 2b = d
S: b = c
let a = 1; c = 1, b = 1 and d = 2
CuCl₂ + H₂S → CuS + 2HCl
A honeybee leaves the hive, flies in a straight line to a flower 6 km away in 15 min, and then takes 15 minutes to return (also in a straight line). a.) Please find the distance travelled and displacement for the entire trip: distance travelled: 1 12 km displacement: 2 0 km b.) Please find the average speed and average velocity for the entire trip: average speed: 3 0.4 km/min average velocity: 4 0.4 km/min c.) If the bee had not flown in a straight line-- but instead with an unknown motion-- to the flower and back, which could not be determined: the average speed or average velocity
Answer:
a) d = 12 km, d = 0, b) v = 0.4 km / min, v = 0, c) v = 0
Explanation:
This problem asks to find the distance and the displacement. We must emphasize that the distance is a scalar and the displacement is a vector quantity
In the exercise, the bold letters indicate vectors
a) Let's find the total distance of the trip
d = d₁ + d₂
where d₁ is the distance to get to the flowers and d₂ is the distance to return to the hive from the flowers
d₁ = d₂ = 6 km
d = 12 km
we look for the displacement that is a vector quantity
d = d₁ - d₂
since the second displacement is in the opposite direction of the first
d = 0
b) the average speed of the trip, again this magnitude is a scalar
v = d / t
distance is d = 12 km t total time is t = 15 + 15 = 30 min
v = 12/30
v = 0.4 km / min
average velocity (vector) of the entire travel, in that case the displacement is zero
v = 0
c) If the path is not a straight line but the time is the same, the average speed cannot be calculated since it lacks the distance value
In the same case to calculate the average velocity that is a vector, it will be zero since the net displacement is zero
v = 0
Question 1 of 10
What might happen to personal information when it is transferred using
digital signals?
A. Some information might be changed when the data are copied.
B. It might be accessed by someone who was not the intended
recipient.
C. The information might change while being transmitted because of
noise.
D. The information might change to analog, making it less reliable.
Answer:
its b for sure
Explanation:
Answer:
B. It might be accessed by someone who was not the intended
recipient