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3 years ago

While eating lunch high up in a skyscraper, two construction workers calculate their gravitational potential

Physics

1 answer:

Maurinko [17]3 years ago

8 0

Answer:

The mass of the other worker is 45 kg

Explanation:

The given parameters are;

The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker

The mass of the lighter construction worker, m₁ = 90 kg

The height level of the lighter construction worker's location = h₁

The height level of the other construction worker's location = h₂ = 2·h₁

The gravitational potential energy, P.E., is given as follows;

P.E. = m·g·h

Where;

m = The mass of the object at height

g = The acceleration due to gravity

h = The height at which is located

Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker

We have;

P.E.₁ = P.E.₂

Therefore;

m₁·g·h₁ = m₂·g·h₂

h₂ = 2·h₁

We have;

m₁·g·h₁ = m₂·g·2·h₁

m₁ = 2·m₂

90 kg = 2 × m₂

m₂ = (90 kg)/2 = 45 kg

The mass of the other construction worker is 45 kg.

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We should first calculate the highest point that ball reaches.
y' = 40 - 32t = 0
t = 40/32 = 1,25s
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To calculate average velocity we use simple formula:
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for t=2 we calculate y
y = 16
(i) for t = 2,5 y = 0
Vav = 16/0.5 = 32 feet/s
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3 years ago

A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy func

OlgaM077 [116]

Answer:

A= 148.92 m/s²

Explanation:

Given that

U(x,y) = (6.00 )x² - (3.75 )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00 )x² - (3.75 )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

- 4.8 = 0.04 x a

a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00 )x² - (3.75 )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

$A=\sqrt{a^2+a'^2}$

$A=\sqrt{120^2+88.2^2}$

A= 148.92 m/s²

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3 years ago