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3 years ago

The total mechanical energy of a basketball is 400 J. If the kinetic energy is 286 J, what must the potential energy be?

Physics

1 answer:

ch4aika [34]3 years ago

7 0

Answer:

the potential energy is 114 J.

Explanation:

Given;

total mechanical energy, E = 400 J

kinetic energy, K.E = 286 J

The potential energy is calculated as follows;

E = K.E + P.E

where;

P.E is the potential energy

P.E = E - K.E

P.E = 400 J - 286 J

P.E = 114 J

Therefore, the potential energy is 114 J.

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A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

3 years ago

I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?

Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

$\frac{1}{2} m {v}^{2} = mgh$

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

$\sin(15) = \frac{h}{d} \\ or \: h = d \sin(15)$

Plugging this into the energy conservation equation and cancelling m, we get

${v}^{2} = 2gd \sin(15)$

Solving for d,

$d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m$

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The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I

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a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

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