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:
(a) 0.833 j
(b) 2.497 j
(c) 4.1625 j
(d) 4.995 watt
Explanation:
We have given force F = 5 N
Mass of the body m = 15 kg
So acceleration 
As the body starts from rest so initial velocity u = 0 m/sec
(a) From second equation of motion 
For t = 1 sec
We know that work done W =force × distance = 5×0.1666 =0.833 j
(b) For t = 2 sec
We know that work done W =force × distance = 5×0.666 =3.33 j
So work done in second second = 3.33-0.833 = 2.497 j
(c) For t = 3 sec
We know that work done W =force × distance = 5×1.4985 =7.4925 j
So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j
(d) Velocity at the end of third second v = u+at
So v = 0+0.333×3 = 0.999 m /sec
We know that power P = force × velocity
So power = 5× 0.999 = 4.995 watt
When you look at the acronym ROY G. BIV, you get the wavelengths from longest to shortest. Since Green is in the middle of ROY G. BIV, anything to the right of it would have a shorter wavelength. You can choose in the visible range Blue, Indigo, or Violet.
<span>A: It is not an exact representation of the atom, but is close enough to be very useful.
Hope this helps!</span>
Answer:
False. the system does not complete the circle movement
Explanation:
For this exercise, we must find the rope tension at the highest point of the path,
-T - W = m a
The acceleration is centripetal
a = v² / R
T = ma - mg
T = m (v² / R - g)
The minimum tension that the rope can have is zero (T = 0)
v² / R - g = 0
v = √ g R
Let's find out what this minimum speed is
v = √ 9.8 1
v = 3.13 m / s
We see that the speed of the body is less than this, so the system does not complete the movement.
