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3 years ago

If you lift one load up two stories, how much work do you do compared to lifting one load up only one story?

Physics

1 answer:

Tatiana [17]3 years ago

7 0

Answer:

D)Twice as much

Explanation:

Work: This is the product of force and distance along the direction of the force. The S. I unit of work is Joules (J)

Energy: Energy is the ability or the capacity to perform work. Energy and work can be interchange during calculation.

E = W = mgh.................................... Equation 1

Where m = mass of the load, h = height of one story, g = acceleration due to gravity.

constant: g = 9.8 m/s²

When the load is lift up only one story

W = 9.8mh..................... Equation 2

But when the load is lifted through two story,

height = 2h

W = 19.6mh .................... Equation 3

Comparing Equation 2 and equation 3,

It shows that the work done to lift the load up two story is twice the work done to lift the load up one story.

The right option is D)Twice as much

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3 years ago

Question 3 (5 points)

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3 years ago

An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul

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Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

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3 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

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3 years ago