The work done by the force in pulling the block all the way to the top of the ramp is 3.486 kJ.
<h3>What is work done?</h3>
Work done is equal to product of force applied and distance moved.
Work = Force x Distance
Given is a block with a weight of 620 N is pulled up at a constant speed on a very smooth ramp by a constant force. The angle of the ramp with respect to the horizontal is θ = 23.5° and the length of the ramp is l = 14.1 m.
From the Newton's law of motion,
ma =F-mg sinθ =0
So, the force F = mg sinθ
Plug the values, we get
F = 620N x sin 23.5°
F = 247.224 N
Work done by motor is W= F x d
The force is equal to the weight F = mg
So, W = 247.224 x 14.1
W = 3.486 kJ
Thus, the work done by the force in pulling the block all the way to the top of the ramp is 3.486 kJ.
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Answer:
Rice
Explanation:
Because I can't control eating lots of rice
Voltage - An electromotive force or potential difference expressed in volts.
Current - A flow of electricity which results from the ordered directional movement of electrically charged particles.
The final velocity of the car is +1.5 m/s.
Explanation:
We have to divide the problem into two parts.
In the first part, the car starts from rest and accelerates for 5.0 s. We can find the final velocity of the car after this first part using the suvat equation:

where
v is the final velocity
u = 0 is the initial velocity
is the acceleration
t = 5.0 s is the time
Substituting,

In the second part, the brakes are applied, so the car decelerates for 3.0 s, and the final velocity is given by

where
v' is the final velocity
v = 7.5 m/s is the velocity at the beginning of this part
is the deceleration
t = 3.0 s is the time
Substituting,

So, the final velocity is +1.5 m/s.
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<span>The runner is moving by uniformly accelerated motion, starting from rest (so, his initial velocity is zero). The law of motion of the runner is
</span>

<span>
where x(t) is the distance covered after time t, and a is the acceleration of the runner. By re-arranging the formula, we get
</span>

<span>
We know the runner has covered a distance of S=12m in t=4.0 s, and if we plug these numbers into the equation, we find the acceleration of the runner:
</span>

<span>
</span>