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djverab [1.8K]
4 years ago
15

Hypothetically speaking, if an object were located at the center of the Earth, the gravitational force on that object due to the

surrounding Earth, assuming matter is uniformly distributed, would have which of the following values?
a) The force would be approximately the same value as if the object were on the surface of the Earth.
b) The force would be much greater than the value if the object were on the surface of the Earth.
c) The force would be somewhat less than the value if the object was on the surface of the Earth, but it would be greater than zero newtons.
d) The force would be zero newtons.
Physics
1 answer:
Likurg_2 [28]4 years ago
5 0

Answer:

D. ) The force would be zero newtons

Explanation:

Because

If you are at the center of the earth, gravity is zero because all the mass around you is pulling "up" (every direction there is up!)

So F=mg so if g is zero F is also zero

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One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor
Bess [88]

Answer:

a) The final velocity is 20 m/s when the large-mass object is the one moving initially.

b) The final velocity is 9.0 m/s when the small-mass object is the one moving initially.

Explanation:

The momentum of the system is calculated as the sum of the momenta of each object. Each momentum is calculated as follows:

p = m · v

Where:

p =  momentum.

m =  mass.

v = velocity.

Then, the momentum of the system is the following:

m1 · v1 + m2 · v2 = (m1 + m2) · v

Where:

m1 = mass of the bigger object.

v1 = velocity of the bigger object.

m2 = mass of the smaller object.

v2 = velocity of the smaller object.

v = final velocity of the two objects after the collision.

Solving the equation for the final velocity:

(m1 · v1 + m2 · v2)/ (m1 + m2) = v

a) Let´s calculate the final velocity when the bigger object is moving:

(7.1 kg · 29 m/s + 3.2 kg · 0)/(7.1 kg + 3.2 kg) = v

<u>v = 20 m/s</u>

b) When the smaller object is moving:

(7.1 kg · 0 m/s + 3.2 kg · 29 m/s) / (7.1 kg + 3.2 kg) = v

<u>v = 9.0 m/s</u>

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3 years ago
A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going
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Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
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After another half-life, 250 are left.
After another half-life, 125 are left.

That was FOUR half-lifes.

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A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

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Answer:

Time period for Simple pendulum, T=2\pi\sqrt{\frac{l}{g}

Explanation:

The Simple Pendulum

Consider a small bob of mass m is tied to extensible string of length l that is fixed to rigid support. The bob is oscillating in the plane about verticle.

       Let \theta is the angle made by string with vertical  during oscillation.

Vertical component of the force on bob, F=-mg\sin\theta

Negative sign shows that its opposing the motion of bob.

Taking \theta as very small angle then, \sin\theta\sim\theta

F=-mg\theta    

Let x is the displacement made by bob from its mean position ,

then, \theta=\frac{x}{l}

so, F=-mg\frac{x}{l}                ........(1)

Since, pendulum is in hormonic motion,

as we know, F=-kx

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\omega=\sqrt{\frac{g}{l}}

Since, \omega=\frac{2\pi}{T}

\frac{2\pi}{T}=\sqrt{\frac{g}{l}

T=2\pi\sqrt{\frac{l}{g}}

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