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Roman55 [17]
3 years ago
12

If a hiker that weighs 600 newtons climbs a 50 meter hill, how much gravitational potential energy has the hiker gained?

Physics
1 answer:
Illusion [34]3 years ago
6 0

The gravitational potential energy U is defined as the product of mass m, the acceleration of gravity g and the height of object h.

U = mgh

We do not have the mass of the hiker. But we know that its W weight is:

W = mg

Where

g = 9.8 m/s^2

So:

m = \frac{W}{g}\\\\m = \frac{600}{g}.

So:

U = (\frac{600}{g})gh\\\\U = (600N)(50m)

U = 30000J

The hiker has gained 30,000 J of energy

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A thin, light wire 75.1 cm long having a circular cross section 0.555 mm in diameter has a 25.4 kg weight attached to it, causin
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Answer:

(a) 3.23×10⁸ N/m²

(b)  1.46×10⁻³

(c) 2.21×10¹¹ N/m²

Explanation:

(a) Stress = Force/Area.

Stress = F/A................ Equation 1

But,

F = mg................. Equation 2

Where m = mass, and g = acceleration due to gravity

A = πd²/4................. Equation 3

d = diameter of the circular cross section.

Substitute equation 2 and equation 3 into equation 1

Stress = 4mg/πd²............. Equation 4

Given: m = 25.4 kg, d = 0.555 mm = 0.000555 m

Constant: g = 9.8 m/s², π = 3.142

Substitute these values into equation 4

Stress = 4(25.4)(9.8)/(3.142×0.00555²)

Stress = 995.68/(3.08×10⁻⁶)

Stress = 3.23×10⁸ N/m²

(b)

Strain = ΔL/L.............. Equation 5

Where ΔL = extension, L = Length.

Given: ΔL = 1.1 mm = 0.0011 m, L = 75.1 cm = 0.751 m

Substitute into equation 5

Strain = 0.0011/0.751

Strain = 1.46×10⁻³.

(c)

Young modulus = Stress/Strain

Young modulus = 3.23×10⁸/ 1.46×10⁻³

Young modulus = 2.21×10¹¹ N/m²

3 0
3 years ago
Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If
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We have: v = d/t
From the expression, we conclude speed is indirectly proportional to speed.
So, the car which will take longer time must have the smallest speed. Among all the options Car C has the smallest speed. So, it would be your answer.

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6 0
3 years ago
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A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a
yawa3891 [41]

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N

- Force B (259 N)

B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N

So the x- and y- component of the total force acting on the block are:

R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N

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3 years ago
What is a disadvantage of using moving water to produce electricity
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A tennis ball is hit at an angle of 20° to the horizontal with the velocity of 20 m/s. What is its velocity in the Y direction?
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B Hopefully I Helped... Good Luck
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