Answer:
39.26 s
Explanation:
From the question given above, the following data were obtainedb
Initial velocity (u) = 6 m/s
Acceleration (a) = 1.9 m/s²
Distance travelled (s) = 1700 m
Time (t) =?
Next, we shall determine the final velocity of the plane. This can be obtained as follow:
Initial velocity (u) = 6 m/s
Acceleration (a) = 1.9 m/s²
Distance travelled (s) = 1700 m
Final velocity (v) =?
v² = u² + 2as
v² = 6² + (2 × 1.9 × 1700)
v² = 36 + 6460
v² = 6496
Take the square root of both side
v = √6496
v = 80.6 m/s
Finally, we shall determine the time taken before the plane lifts off. This can be obtained as follow:
Initial velocity (u) = 6 m/s
Acceleration (a) = 1.9 m/s²
Final velocity (v) = 80.6 m/s
Time (t) =?
v = u + at
80.6 = 6 + 1.9t
Collect like terms
80.6 – 6 = 1.9t
74.6 = 1.9t
Divide both side by 1.9
t = 74.6 / 1.9
t = 39.26 s
This, it will take 39.26 s before the plane lifts off.
calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)
A 4.0-kilogram ball held 1.5 meters above the floor has ________ joules of potential energy
Answer:
58.8J
Explanation:
Given parameters;
Mass of ball = 4kg
Height above the floor = 1.5m
g = 9.8n/kg
Unknown:
Potential energy = ?
Solution:
The potential energy of a body is the energy due to the position of the body.
It is mathematically expressed as:
Potential energy = mass x acceleration due to gravity x height
Potential energy = 4 x 9.8 x 1.5 = 58.8J
if you watch football let me know who you think is going to win super bowl 55 and what do you think the score going to be Kansas city chiefs or tampa bay buccaneers
Answer:
I think the bucs are gonna win because Tom Brady is on their team and it's rigged
but maybe I'm just thinking negatively lol
Which of the following is an action-at-a-distance force? friction tension gravity air resistance
Answer:
Action-at-a-Distance Forces. Frictional Force. Gravitational Force. Tension Force ... The force of gravity on earth is always equal to the weight of the object as ... The friction force is the force exerted by a surface as an object moves across it or ... The force of air resistance is often observed to oppose the motion of an object
Explanation:
Answer:
I believe the answer is gravity.
Explanation:
Because with friction you need to be rubbing multiple object together, in tension two objects must be pulling against each other, and in air resistance the air must be touching the object either pushing or pulling it. While in gravity the mass of an object is pulling another object toward it, the objects don't have to be touching each other making it an at-a-distance force.
You are moving a wagon with a friend's help you push on the left side of the wagon with 25 of force while your friend pulls from the right side of the wagon with a force of 15N,
What is the net force?
Answer:
10N to the left side towards you
Explanation:
The net force is the resultant force that acts on a body.
Force is a push or pull on a body.
Push to left side = 25N
Pull to the right = 15N
Net force = Push to left side - Pull to the right = 25N - 15N
Net force = 10N to the left side towards you
The net force is therefore 10N to the left side towards you
A chocolate chip cookie is an example of a (2 points) a homogeneous mixture b heterogeneous mixture c suspension d colloid
Answer:
I think it is heterogeneous mixture. have a good day
Answer:
heterogeneous mixture
Explanation:
i took the test
What Coulombs discovered almost 300
years ago
Answer:
ummm hehe this is my time to shine
Explanation:
MERICIA!!!!!!!!!!!!!!!!!!!!!!!
Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer distance
Please show working
Distance = (speed) x (time)
Car A: Distance = (8 m/s) x (43 s) = 344 meters
Car B: Distance = (7 m/s) x (50 s) = 350 meters
350 meters is a longer distance than 344 meters.
Car-B traveled a longer distance than Car-A did.
Answer:
[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]
Explanation:
Distance is equal to the product of speed and time.
[tex]d=s*t[/tex]
1. Car A
Car A has a speed of 8 meters per second and travels for 43 seconds.
[tex]s= 8 \ m/s \\t= 43 \ s[/tex]
Substitute the values into the formula.
[tex]d= 8 \ m/s *43 \ s[/tex]
Multiply and note that the seconds will cancel out.
[tex]d= 8 \ m*43= 344 \ m[/tex]
2. Car B
Car B has a speed of 7 meters per second and travels for 50 seconds.
[tex]s= 7 \ m/s \\t= 50 \ s[/tex]
Substitute the values in and multiply.
[tex]d= 7 \ m/s * 50 \ s[/tex]
[tex]d= 7 \ m * 50 = 350 \ m[/tex]
350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.
I need help with questions b and d, that’s all.
Thank you.
b). The power depends on the RATE at which work is done.
Power = (Work or Energy) / (time)
So to calculate it, you have to know how much work is done AND how much time that takes.
In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.
The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.
Without knowing the time, we can't calculate the power.
...
d). Kinetic energy = (1/2) · (mass) · (speed squared)
On the way up, the car stops when it reaches point-A.
On the way down, the car leaves point-A from "rest".
WHILE it's at point-A, it has no speed. So it has no (zero) kinetic energy.
which type of electromagnetic radiation is most likely to cause skin cancer as a result of sun exposure over time?
Answer:
Ultraviolet radiation
Explanation:
The type of electromagnetic radiation most likely to cause skin cancer as a result of sun exposure overtime is the ultraviolet radiation.
Electromagnetic radiation occurs with a broad spectrum starting from gamma rays to the radio waves.
From one end to the other, their energy decreases as the wavelength increases.
Within this broad spectrum, the ultraviolet rays which are before the visible rays are very energetic and can cause skin cancer.
Answer:
ultraviolet i think sorry if im wrong :/
Explanation:
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave perspective, the antenna is a circular aperture through which the microwaves diffract. What is the diameter of the radar beam at a distance of 30 km
Answer:
915m
Hope this helps.
A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to 14C in 6.35 s. The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of 1.263 kg. If the same observed temperature change occurs in 4.59 s, what is the specific heat of the unknown material
Answer:
The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Explanation:
Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:
[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the sphere, measured in kilograms.
[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.
[tex]\Delta t[/tex] - Time, measured in seconds.
In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:
[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)
Where:
[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.
[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.
[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.
[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]
If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:
[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]
[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.1 ss for the boat to travel from its highest point to its lowest, a total distance of 0.55 mm. The fisherman sees that the wave crests are spaced 4.2 mm apart.
A. How fast are the waves traveling?
B. What is the amplitude of each wave?
C. If the total vertical distance traveled by the boat were 0.500, but the other data remained the same, how fast are the waves traveling?
D. If the total vertical distance traveled by the boat were 0.500, but the other data remained the same, what is the amplitude of each wave?
Answer:
Explanation:
It takes 3.1 s for the boat to travel from its highest point to its lowest, so the period of oscillation
T = 2 x 3.1 = 6.2 s
frequency of wave n = 1 / T = .1613 per sec
Amplitude of oscillation = .55/2 = .275 mm
The fisherman sees that the wave crests are spaced 4.2 mm apart. so wavelength of wave λ = 4.2 mm .
A ) velocity of wave v = n λ
.1613 x 4.2 = .677 mm /s
B ) Amplitude of wave = .275 mm
C ) The vertical distance determines only the amplitude which does not affect the velocity , so velocity will remain unchanged .
D ) Amplitude of wave depends only on the vertical displacement .
The amplitude will become .5 / 2 = .25 mm .
A cylindrical tank of radius R, filled to the top with a liquid, has a small hole in the side, of radius r, at distance d below the surface. Find an expression for the volume flow rate through the hole. A 4.0-mm-diameter hole is 1.0 m below the surface of a 2.0-m-diameter tank of water. What is the rate, in mm/min, at which the water level will initially drop if the water is not replenished?
Answer:
a)[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
b)[tex]dh / dt = 0.2658 mm / min[/tex]
Explanation:
From the question we are told that
Diameter of hole [tex]d_h=4mm=>0.004m[/tex]
Depth of hole [tex]D=0mm=>0.001m[/tex]
Diameter of tank [tex]d_t=2mm=>0.002m[/tex]
Generally the equation for pressure is mathematically given as
[tex]Pressure P= \rho*g*d[/tex]
[tex]P= 1/2*\rho *v^2[/tex]
Where
[tex]v = \sqrt {2gd}[/tex]
[tex]V = Area*v[/tex]
[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by
[tex]dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s \\[/tex]
Therefore the level at which the water level will initially drop if the water is not replenished
[tex]dh / dt = 0.2658 mm / min[/tex]
The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Given data:
The diameter of hole is, d = 4.0 mm = 0.004 m.
The depth of hole is, h = 1.0 m.
The diameter of tank is, d' = 2.0 m.
The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.
Let us first obtain the equation of pressure as,
[tex]P=\dfrac{1}{2} \times \rho \times v^{2}[/tex]
Here, v is the velocity of efflux and its value is,
[tex]v=\sqrt{2gh} \\\\v^{2}=2gh[/tex]
And the level at which the water level will initially drop if the water is not replenished is mathematically given by,
[tex]\dfrac{dH}{dt}=(r/R)^{2} \times v[/tex]
Here,
r is the radius of hole.
R is the radius of tank.
Solving as,
[tex]\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min[/tex]
Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Learn more about the flow rate here:
https://brainly.com/question/11816739
Required
Momentum
The magnitude of the momentum of an object is 64 kg*m/s. If the velocity of the
object is doubled, what will be the magnitude of the momentum of the object? *
32 kg*m/s
64 kg*m/s
128 kg*m/s
256 kg*m/s
Answer:
C) 128 kg*m/s
Explanation:
When you double something you multiply it by 2 most of the time. 64*2=128 or you can add it 64+64=128. Hope this helps.
1. What average force is exerted on a 25 g egg by a bed sheet if the egg hits the sheet at 4 m/s and takes
0.2 s to stop?
Answer:
F = -0.5 N
Explanation:
Given that,
Mass of an egg, m = 25 g = 0.025 kg
Initial speed, u = 4 m/s
Final speed, v = 0 (it stops)
Time, t = 0.2 s
We need to find the average force exerted on the egg. The force is given by :
F = ma
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.025\times (0-4)}{0.2}\\\\F=-0.5\ N[/tex]
So, the average force exerted on the egg is (-0.5 N).
A hazard sign has 3 identical
parallelogram-shaped stripes as shown.
Charles must outline each stripe with
reflective tape. Is one roll of 144 inches
of tape enough to finish the job?
Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.
Perimeter is the sum of all the sides of a figure.
For a parallelogram:
P = 2*length + 2*width
So, we need to determine width and length of the stripe.
Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is
[tex]h=\sqrt{18^{2}+6^{2}}[/tex]
[tex]h=\sqrt{360}[/tex]
h = 19 in
Perimeter of the first stripe is
P = (2*19) + (2*3)
P = 44 inches
The hazard sign has 3 stripes. So total perimeter is
[tex]P_{t}=[/tex] 44 + 44 + 44
[tex]P_{t}=[/tex] 132 inches
To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.
Water flows through a first pipe of diameter 3 inches. If it is desired to use another pipe for the same flow rate such that the velocity head in the second pipe is twice the velocity head in the first pipe, determine the diameter of the second pipe.
Answer:
the diameter of the second pipe is 2.52 in
Explanation:
Given the data in the question;
We know that; the rate of flow is the same;
so
Av1 = Av2
v ∝ √h
[tex]\frac{A1}{A2}[/tex] = [tex]\frac{V2}{V1}[/tex]
[tex]\frac{A1}{A2}[/tex] = √( [tex]\frac{h2}{h1}[/tex] )
( π/4.D1² / π/4.D2² ) = √( [tex]\frac{h2}{h1}[/tex] )
( D1² / D2² ) = √( [tex]\frac{2h1}{h1}[/tex] ) since second is double of first
so
( D1² / D2² ) = √( [tex]\frac{2}{1}[/tex] )
3² / D2² = √2
D2²√2 = 9
D2² = 9/√2
D2² = 9 / 1.4142
D2² = 6.364
D2 = √ 6.364
D2 = 2.52 in
Therefore, the diameter of the second pipe is 2.52 in
The process of braking or accelerating an automobile in heavy traffic is highly complex, requiring the skillful use of both feedback and feedforward mechanisms to drive safely. Consider a controlled variable as the distance between you and the car in front of you, with some specified distance as a set point. Name one feedback and one feedforward control mechanisms would keep you at that distance, and prevent you from colliding with the car in front of you
Answer:
A feedback mechanism is the constant measurement of the distance between the two vehicles and with this the calculation of the speed between them
An anticipated control mechanism is using the vehicle's acceleration and its deceleration to calculate the future speed and their distances,
Explanation:
For this exercise, the feedback and control mechanisms must be directly related to the kinematic relationships,
It is assumed that the vehicle speed (taken from the speedometer) and the braking capacity (given by the brake manufacturer) are known in the form of negative acceleration,
A feedback mechanism is the constant measurement of the distance between the two vehicles and with this the calculation of the speed between them, for which we know the acceleration that exists. This would be a correct mechanism, in general we can adjust to an error between the sedated distance and the real one, so when they have very different give a maximum acceleration and in decreasing it as the differences between the distances decrease.
An anticipated control mechanism is using the vehicle's acceleration and its deceleration to calculate the future speed and their distances, so we would know the amount of acceleration necessary to reach the optimal distance between the two vehicles.
F=9 N, a=3 m/s², m=?
Answer:
3kg
Explanation:
Given parameters:
Force = 9N
Acceleration = 3m/s²
Unknown:
Mass = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
So;
9 = mass x 3
mass = 3kg
A student is measuring the mass of 20 paper clips using an electronic balance that measures to the thousandths of a gram. The balance displays the value for the mass of the paper clips: 20.120 g. Which of the values would be acceptable ways to record this mass in a lab notebook
The question is incomplete, the complete question is;
A student is measuring the mass of 20 paper clips using an electric balance that measures to the thousandths of a gram. The balance displays the following values for the mass of the paper clip: 20.120 g.
Which of the following choices would be acceptable ways to record this mass on a lab report? Select all that apply.
a) 20g
b) 20.1 g
c) 20.12 g
d) 20.120g
e) 20.1200g
Answer:
d) 20.120g
Explanation:
We have been told in the question that the electronic balance measures the mass of the paper clips to thousandths of a gram.
This implies that the value of mass is measured to the third decimal place.
If we look at the options, 20.120 g is the measurement of the mass to thousandths of a gram hence that is the correct answer to the question.
A particle has a velocity that is 90.% of the speed of light. If the wavelength of the particle is 1.5 x 10^-15 m, calculate the mass of the particle
Answer:
[tex]m=1.63\times 10^{-27}\ kg[/tex]
Explanation:
The velocity of a particle is 90% of the speed of light.
The wavelength of the particle is [tex]1.5\times 10^{-15}\ m[/tex]
We need to find the mass of the particle.
The formula for the wavelength of a particle is given by :
[tex]\lambda=\dfrac{h}{mv}[/tex]
h is Planck's constant
v is 90% of speed of light
m is mass of the particle
[tex]m=\dfrac{h}{\lambda v}\\\\m=\dfrac{6.63\times 10^{-34}}{1.5\times 10^{-15}\times 0.9\times 3\times 10^8}\\\\m=1.63\times 10^{-27}\ kg[/tex]
So, the mass of the particle is [tex]1.63\times 10^{-27}\ kg[/tex].
25 points!
A 6 kg object accelerates from 5 m•s to 25 m•s in 30 seconds. What was the net force acting on the
object? Give your answer in Newtons to one significant figure and without a unit.
(Show Work)
Answer:
6N
Explanation:
Given parameters:
Mass of object = 6kg
Initial velocity = 5m/s
Final velocity = 25m/s
Time = 30s
Unknown:
Net force acting on the object = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration is the rate of change of velocity with time
Acceleration = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
Force = mass x [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
So;
Force = 6 x [tex]\frac{25 - 5}{30}[/tex] = 6N
3. A 200-g ball falls vertically downward, hitting the floor with a speed of 2.5 m/s and rebounding upward with a speed of 2.0 m/s. a. Determine the change in momentum of the ball.. b. If the ball is in contact with the floor for 0.02 ms (milliseconds), what is the average force applied to the ball
Answer:
Explanation:
A 200-g ball falls vertically downward, hitting the floor with a speed of 2.5 m/s and rebounding upward with a speed of 2.0 m/s. a. Determine the change in momentum of the ball.. b. If the ball is in contact with the floor for 0.02 ms (milliseconds), what is the average force applied to the ball
Given data
mass= 200g= 0.2kg
initial velocity= 2.5m/s
final velocity= 2m/s
time= 0.02ms
time= 0.00002 seconds
ΔP= mΔv
ΔP= 0.2*2.5-2
ΔP= 0.2*0.5
ΔP=0.1kgm/s
F= mv/t
F=0.1/0.00002
F=5000N
Which of the following do not involve a direction?
Check all that apply.
A. Velocity
B. Distance
C. Time
D. Acceleration
Distance and time I think
why do players choose to follow the unconventional route of kicking down the middle
Answer:
My biggest reason is to make it a habit. Even if the ball goes into the endzone it is a live ball and the offensive players must down the ball. Don't leave any room for "I thought he downed it" or "I thought I heard the whistle" just run to the ball always.
If the players slow down and the returner takes it out of the end zone it could be a big return. Players are on a full sprint for 40+ yards sometimes and instead of breaking down, they choose to contine through the goal line to slow down at a decreased rate (possibly limiting a muscle pull injury).
what is the weight in Newtons of a ball with a mass of 7.77 kg?
Answer:
76.1N
Explanation:
Given parameters:
Mass of the ball = 7.77kg
Unknow:
Weight of balloon = ?
Solution:
Weight is the vertical force applied on a body.
Weight = mass x acceleration due gravity
So;
Weight = mass x acceleration due to gravity
So;
Weight = 7.77 x 9.8 = 76.1N
When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2 . What displacement does it undergo during this deceleration period
Answer:
240 meters
Explanation:
The distance traveled by the vehicle can be calculated using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex] (1)
Where:
x: is the displacement
[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)
[tex]v_{0}[/tex]: is the initial speed = 60 m/s
a: is the acceleration = -7.5 m/s²
By solving equation (1) for x we have:
[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]
Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.
I hope it helps you!
It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 390 m the field has magnitude 60.0 N/C. At an altitude of 240 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 150 m on edge, with horizontal faces at altitudes of 240 and 390 m. Neglect the curvature of Earth.
Answer:
[tex]q=7.965*10^-^6C[/tex]
Explanation:
From the question we are told that
Altitude of [tex]d_1 390m[/tex]
Magnitude[tex]M_1=60.0 N/C[/tex]
Altitude of [tex]d_2=240 m[/tex]
Magnitude is [tex]M_2= 100 N/C[/tex]
Distance of cube [tex]d_c=150 m[/tex]
Generally the flux [tex]\phi[/tex] is mathematical given as
[tex]\phi=60(150)^2cos180+100(150)^2*cos0[/tex]
[tex]\phi=-9*10^5[/tex]
Generally Quantity of charge q is mathematically given as
[tex]q=\varepsilon _0 *\phi[/tex]
[tex]q=8.85*10^-^1^2 *9*10^5[/tex]
[tex]q=7.965*10^-^6C[/tex]
Which of the following has kinetic energy? (1 point)
Ans:
C
Explanation:
because it is moving down the stairs
A spring travelling down a flight of stairs has kinetic energy.
What is Kinetic energy?
Since it is moving, it would be a spring moving down a set of steps. A stretched spring, a compressed spring, and a spring at the top of a set of steps are all in motion.
Kinetic energy is a type of power that a moving object or particle possesses.
An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force. A moving object or particle has kinetic energy, which depends on both its mass and its rate of motion.
Therefore, A spring travelling down a flight of stairs has kinetic energy.
To learn more about Kinetic energy, refer to the link:
https://brainly.com/question/26472013
#SPJ5
The pressure at the bottom of a cylindrical container with a cross-sectional area of 45.5 cm2 and holding a fluid of density 420 kg/m3 is 115 kPa. (a) Determine the depth of the fluid. How is the pressure on the bottom of the container related to atmospheric pressure and the pressure due to the depth of the fluid
Answer:
3.33 m
Explanation:
Pressure is the distributed force applied to the surface of an object per unit area. The force is applied perpendicular to the surface of the object. The SI unit of pressure is N/m² or Pa.
Hydrostatic pressure is the pressure that a fluid exerts at a point due to the force of gravity.
The relationship between pressure on the bottom of the container, atmospheric pressure and the pressure due to the depth of the fluid is given by:
[tex]P_{bottom}-P_{atm}=P_{depth}\\\\where\ P_{bottom}=pressure\ at\ the \ fluid\ bottom,\ P_{atm}=atmospheric\ pressure\\P_{depth}=pressure\ due\ to\ fluid\ depth=\rho gh. \ Hence:\\\\P_{bottom}-P_{atm}=\rho gh\\\\Given \ that\ P_{bottom}=115\ kPa=115*10^3\ Pa, let\ us\ assume\ P_{atm}=101\ kPa=101*10^3\ Pa,\rho=420\ kg/m^3,g=acceleration\ due\ to \ gravity=10\ m/s^2.\\\\Therefore:\\\\115*10^3-101*10^3=420*10*h\\\\14*10^3=4200h\\\\h=3.33\ m\\\\[/tex]