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defon
3 years ago
5

What is the reaction force if a girl pulls on a cow?

Physics
1 answer:
Papessa [141]3 years ago
8 0
Answer : B) The cow pulls back on the girl.

From newton’s third law we know that every action has a reaction force pushing back. So when the girl pulls on a cow, the cow is pulling back on her.
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An electrically neutral atom has the same # of protons and electrons true or false ?
Airida [17]

Answer:

True

Explanation:

A neutral atom has same no. of protons and electron that's why they are neutral.

5 0
3 years ago
Read 2 more answers
A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.36
Semmy [17]

Answer: the average speed of the rat from the information given above is 0.7m/s

Explanation:

position is given as

x(t) = pt² + qt

finding the diffencial of x(t) with respect to t, we have

d(x(t))/dt = 2pt + q

we substitute the p = 0.36m/s² and q= -1.10 m/s

d(x(t))/dt = 2(0.36)t + (-1.10)

so, at t= 1s

d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s

at t= 4s

d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s

To find the average speed,

average speed = (V1 + V2)/ 2

average speed = (1.78 + (-0.38))/2 = 0.7m/s

5 0
3 years ago
A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle
Alex73 [517]

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

6 0
3 years ago
Que la tarcer ley de newton
slava [35]

Explanation:

If you write it in English so I can help u if you need it

3 0
3 years ago
26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the
denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

U=mg \Delta h

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

\Delta h = 4 m + 0.02 m = 4.02 m is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J

4 0
3 years ago
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