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Anna [14]
3 years ago
12

What is the momentum of a 1200 kg car traveling with a speed of 27 m/s (60 mph)?

Physics
1 answer:
serious [3.7K]3 years ago
3 0

Answer:

This is your answer

Explanation:

Actually I took this from go ogle

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The theory that the outer core of the Earth is
ELEN [110]
Scientists can tell this by looking at seismic waves released by earthquakes. ( if you need me to elaborate more ask)
4 0
3 years ago
What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?
yKpoI14uk [10]

Answer:

<h3>The answer is 5.4 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{135}{25}   =  \frac{27}{5} \\

We have the final answer as

<h3>5.4 kg</h3>

Hope this helps you

4 0
3 years ago
Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3
RideAnS [48]

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 12.6 N

d = 15.4 m

\theta = 15.3 degree

so we will have

W = 12.6 \times 15.4 cos15.3

W = 187.2 J

5 0
3 years ago
The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra
Scrat [10]

Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

7 0
3 years ago
Why is there zero current when a light bulb burns out?
kozerog [31]

The circuit is no longer closed.

8 0
4 years ago
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