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3 years ago

a 2.80 kg mass is dropped from a height of 4.50 m. find its potential energy(PE) at the moment it is dropped. PLEASE HELP

Physics

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

126 J

Explanation:

The potential energy only depends on the vertical height from the ground level.

We consider the ground level to have zero P.E.

We get, when it is 4.5 m above the ground level,

P.E. = mgh

= 2.8×10×4.5

= 126 J

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Explanation:

Given

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3 years ago

A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo

Drupady [299]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

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2 years ago

What would happen to a moving object if no unbalanced force acts upon it?<br> Help please?

Kisachek [45]

Answer:

Explanation:

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hope this helps!! have a wonderfull day!

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2 years ago

D. When Alan's puck is moving at 21 m/s south, it collides

kirza4 [7]

D is s s s s s ss s s s s

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2 years ago

A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th

ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

$T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin$

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

$\frac{16T_{1} }{\pi d_{1}^{3} } =\frac{16T_{2} }{\pi d_{2}^{3} }$ (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4} }\\T_{1} =16216d_{1} ^{4}$

The same way to θ₂:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4} } \\T_{2} =9523.8d_{2} ^{4}$

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

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2 years ago