The frictional force is 39.4 N
Explanation:
We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

where
is the net force
m is the mass
a is the acceleration
Here we know that the box is moving with constant velocity, so its acceleration is zero:

This means that the net force is also zero:

The net force on the block is given by the applied force, forward, and the frictional force, backward:

where
is the applied force
is the frictional force
Therefore, solving for
,

Learn more about friction:
brainly.com/question/6217246
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<span>Answer a. is correct. In this case the dog treats were positive reinforcers for staying close (the desirable behaviour). Repeated rewards for a certain behaviour work as positive reinforcers.</span>
Answer:
v' = 2.4 m/s
Explanation:
Given that,
Mass of one skater, m = 60 kg
Mass of the other's skater, m' = 60 kg
The two skaters push off each other. After the push, the smaller skater has a velocity of 3.0 m/s.
When there is no external force acting on a system, the momentum remains conserved. It means initial momentum is equal to the final momentum. Let v' is the velocity of the larger skater.
mv = m'v'

So, the velocity of the larger skater is 2.4 m/s.
Answer:
Gravity on the moon, g = 1.69 m/s²
Explanation:
It is given that,
Length of pendulum, l = 1 m
Time period, T = 4.82 seconds
We have to find the gravity of the moon. The time period of the pendulum is given by :

g = acceleration due to gravity on moon


g = 1.69 m/s²
Hence, the gravity on the moon is 1.69 m/s².
Answer:
i think c)light has a charge