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3 years ago

3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the

Physics

1 answer:

andreev551 [17]3 years ago

5 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

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At the instant shown in the diagram, the car's centripetal acceleration is directed

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Answer:

At the instant shown in the diagram, the car's centripetal acceleration is directed is discussed below in detail.

Explanation:

The direction of the centripetal acceleration is in a circular movement is forever towards the middle of the roundabout pathway. In the picture displayed, the East direction is approaching the center. So, the course of the car's centripetal acceleration is (H) toward the east.

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3 years ago

A ball is rolled twice across the same level laboratory table and allowed to roll off the table and strike the floor. In each tr

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By definition we know that the distance is equal to the speed by time
d = v * t
Clearing the time we have
t = d / v
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Clearing the speed
v = Root (2gh)
Then, substituting
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We conclude that the time is the same since it depends on the height of the table to the floor.

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3 years ago

What is the meaning of electrolysis

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Answer:

Electrolysis is the passing of a direct electric current through an ionic substance that is either molten or dissolved in a suitable solvent, producing chemical reactions at the electrodes and decomposition of the materials.

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3 years ago

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A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the hor

Wittaler [7]

Answer:

7.78 meters

Explanation:

So, lets just pretend we are working in a vacuum, like in (almost) all physics problems. So, we must find how much time will take to the man fall the height to the saddle to know how far away must be the saddle to be right under the limb at the right time.

So, we can use kinematics relationships. We know the for a 1D movement, with constant acceleration, the equation its:

$y(t) \ = \ y_0 \ + \ v_0 \ * \ t \ +\frac{1}{2} \ a \ t^2$

For our problem, the initial position its the height from which the man jumps,

$y_o \ = 2.45 \m$ , the final position

$y(t_{fall})$ must be the level of the saddle

$y(t_{fall}) \ = 0 \ m$ ,

the initial velocity must be

$v_0 \ = \ 0 \ \frac{m}{s}$ ,

and the acceleration its the gravitational pull,

$a \ = \ - \ g \ = \ - \ 9.8 \ \frac{m}{s^2}$ .

So, we get:

$0\ = \ 2.45 \ m \ + \ 0 \ * \ t \ - \frac{1}{2} \ g \ t_{fall}^2$

$0\ = \ 2.45 \ m \ - \frac{1}{2} \ g \ t_{fall}^2$

$- \ 2.45 \ m \ = - \frac{1}{2} \ g \ t_{fall}^2$

$\ 2.45 \ m \ = \frac{1}{2} \ g \ t_{fall}^2$

$2 \ * \ 2.45 \ m \ = \ g \ t_{fall}^2$

$\frac{ 2 \ * \ 2.45 \ m}{g} \ = t_{fall}^2$

$\sqrt{ \frac{ 2 \ * \ 2.45 \ m}{g} } \ = \ t_{fall}$

So, we can calculate this an get

$t_{fall} \ = \sqrt{ \frac{ 2 \ * \ 2.45 \ m}{9.8 \frac{m}{s^2} } }$

$t_{fall} \ = \sqrt{ \frac{1}{2} }$

Now, we know how much time will take for the man to fall to the level of the saddle. If the horse is galloping to a constant speed of

$v_h \ = \ 11.0 \ \frac{m}{s}$ ,

in the time $t__{fall}$ the horse travels a distance

$d_h \ = \ v_h * t_{fall}$

$d_h \ = \ 11.0 \ \frac{m}{s} * \sqrt{ \frac{1}{2} }$

$d_h \ = \ 7.78 \ m$

And this must be the distance we are looking for. So, the saddle and the limb must be at 7.78 meters when the man makes his move.

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3 years ago

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Reika [66]

Answer:

yess

Explanation:

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3 years ago