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alexdok [17]
3 years ago
8

Which of the following properties could be described as a force that pushes charges along?

Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0
C) voltage is a force that pushes charge along
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What is the first law of thermodynamics?​
jok3333 [9.3K]

Answer:

it states that energy can neither be created or destroyed

4 0
3 years ago
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A boy in a wheelchair (total mass 54.5 kg) has speed 1.40 m/s at the crest of a slope 2.10 m high and 12.4 m long. At the bottom
babymother [125]

Answer:

630.75 j

Explanation:

from the question we have the following

total mass (m) = 54.5 kg

initial speed (Vi) = 1.4 m/s

final speed (Vf) = 6.6 m/s

frictional force (FF) = 41 N

height of slope (h) = 2.1 m

length of slope (d) = 12.4 m

acceleration due to gravity (g) = 9.8 m/s^2

work done (wd) = ?

  • we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy

wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d)

wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d) - (mgh)

where wd = work done

m = mass

h = height

g = acceleration due to gravity

FF = frictional force

d = distance

Vf and Vi = final and initial velocity

wd =  (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)            

wd = 630.75 j

3 0
3 years ago
The formula is x = 1/2 at^2 and I have managed to fill in the variables as this. d = 1/2 9.81 m/s^2 1^2
Artyom0805 [142]

Right, as you mentioned in the comments, you find d by plugging in the different values of t.

For t=1\,\mathrm s, we have

d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1\,\mathrm s)^2

d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(1\,\mathrm s^2\right)

d=4.905\,\mathrm m

Similarly, for t=2\,\mathrm s, you get

d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(2\,\mathrm s\right)

d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(4\,\mathrm s^2\right)

d=19.62\,\mathrm m

8 0
3 years ago
A runaway railroad car, with mass 30x10^4 kg, coasts across a level track at 2.0 m/s when it collides with a spring loaded bumpe
Natalija [7]

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2

where we have

k=2\cdot 10^6 N/m is the spring constant of the bumper

x is the maximum compression of the bumper

m=30\cdot 10^4 kg is the mass of the car

v=2.0 m/s is the speed of the car

Solving for x, we find the maximum compression of the spring:

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m

8 0
3 years ago
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What did copernicus say about the motion of the sun?.
eimsori [14]

the sun was stationary in the center of the universe and the earth revolved around it....

8 0
2 years ago
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