Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end
of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2750 2750 Newtons while Magnus pulls the bus a distance of 1.60 1.60 meters, how much work does the tension force do on Magnus? The rope is perfectly horizontal during the pull.
1 answer:
Answer:
The workdone is
Explanation:
The free body diagram is shown on the first uploaded image
From the question we are given that
The force is on the force gauge
The distance that Magnus pulled the bus
Generally the workdone by the tension force on Magnus is
This negative sign show that is tension force is in the opposite direction to Magnus movement (i.e the movement of the bus )
Substituting value we have
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Answer:
v_wind = 101.46 km / h , θ = 61.8
Explanation:
This is a velocity composition exercise.
Let's do the problem in parts. Let's start by knowing the speed of the plane without air.
v = d / t
v = 750 / 3.14
v = 238.85 km / h
This is the speed of the plane relative to the Earth and it does not change.
In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and
Let's use trigonometry to find the components of the plane's speed
cos θ = v_N / v
sin θ = v_W / v
v_N = v cos θ
v_W = v sin θ
let's calculate
V _N = 238.85 cos 22 = 221.46 km / h
v_W = -238.85 sin 22 = -89.47
the negative sign is because the plane is going west and the positive sign is the east direction.
As it indicates that the destination of the avine is towards the north, the x component of the wind must be
vₓ - v_W = 0
vₓ = v-w
vₓ = 89.47 km / h
in the direction to the East.
Now let's analyze the component of the wind in the Nort-South direction,
Indicate the travel time, let's calculate the speed that the component must have the speed of the plane
v_total = d / t
v_total = 750 / 4.32
v_total = 173.61 km / h
This is the final speed of the plane, which can be written
v_total = v_n - vy
vy = v_n - v_total
vy = 221.46 - 173.61
vy = 47.85 km
this component is directed towards the south
Let's use the Pythagorean Theorem, to find the magnitude
v_wind² = vₓ² + vy²
v_wind = √ (89.47² + 47.85²)
v_wind = 101.46 km / h
the address will then be found using trigonometry
θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹1 (47.85 / 89.47)
θ = 28.14
Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value
90- θtea = 90- 28.2
θ = 61.8
East of South
Okay, haven't done physics in years, let's see if I remember this.
So Coulomb's Law states that
so if we double the charge on 
and double the distance to 
we plug these into the equation to find
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So we see the new force is exactly 1/2 of the old force so your answer should be
if I can remember my physics correctly.
So Coulomb's Law states that
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So we see the new force is exactly 1/2 of the old force so your answer should be