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3 years ago

This graph shows the change in the force of gravity as distance increases. The graph will never reach zero, which suggests that

Physics

1 answer:

kumpel [21]3 years ago

7 0

Answer:

Gravity can never be zero.

Explanation:

The graph shows that the force of gravity is always present despite the amount of distance, indicating that gravity will never be 0. Note, however, that gravity can be extremely small as the distance increases. The limit of the force of gravity as the distance approaches infinity is 0, but never actually touches this value.

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gizmo_the_mogwai [7]

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2 years ago

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Ede4ka [16]

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2 years ago

After t = 0.38 s what is the speed of the block?

scoray [572]

$v = \frac{dx}{dt} = \int\limits^{t_2}_{t_1} {a} \, dt$

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3 years ago

A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin

Goshia [24]

Answer:

(A) 0.63 J

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + m $h^{2}$

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = $(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}$ )^{2}[/tex]

I = $(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}$ )^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5I $ω^{2}$

= 0.5 x 0.07875 x $4^{2}$ = 0.63 J 0.15 m

(B) from the conservation of energy

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

Ki + Ui = Kf + Uf

at the maximum height velocity = 0 therefore final kinetic energy = 0

Ki + Ui = Uf

Ki = Uf - Ui

Ki = mg(H-h)

where (H-h) = rise in the center of mass

0.63 = 0.42 x 9.8 x (H-h)

(H-h) = 0.15 m

6 0

3 years ago

A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot

Reil [10]

To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
= 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
= 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2

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3 years ago

This graph shows the change in the force of gravity as distance increases. The graph will never reach zero, which suggests that​

This graph shows the change in the force of gravity as distance increases. The graph will never reach zero, which suggests that