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Elina [12.6K]
3 years ago
12

Batman, whose mass is 89.5 kg, is holding on to the free end of a 13.4 m rope, the other end of which is fixed to a tree limb ab

ove. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough that he can reach a ledge when the rope makes a 72.4 ◦ angle with the vertical. The acceleration of gravity is 9.8 m/s 2 . How much work was done against the force of gravity in this maneuver?
Physics
1 answer:
astra-53 [7]3 years ago
3 0

Answer:

8.2\times 10^{3} J

Explanation:

L = length of the rope connected = 13.4 m

\theta = Angle made by the rope with the vertical = 72.4°

h = height gained by the batman above the reference point  

Height gained by the batman above the reference point  is given as

h = L(1 - Cos\theta )

m = mass of batman = 89.5 kg

g = acceleration due to gravity = 9.8 ms⁻²

Work done against gravity is same as the gravitational potential energy gained by batman and is given as

W = mgh

W = mgL(1 - Cos\theta )

Inserting the above values

W = (89.5)(9.8)(13.4)(1 - Cos72.4 )

W = 8.2\times 10^{3} J

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3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

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4 years ago
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Nady [450]

Answer:

Any Lens

Explanation:

I Hope it's right if not so Sorry :)

3 0
3 years ago
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Licemer1 [7]

Answer:

<h3>30m</h3>

Explanation:

Velocity is the change of rate of displacement with respect to time.

velocity = displacement/time

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initial velocity = 15 m/s.

time taken =2 secs

Required

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From the formula;

Displacement = Velocity * time

Displacement = 15 * 2

Displacement = 30m

<em>Hence the displacement of the object is 30m</em>

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7 0
3 years ago
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