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2 years ago

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A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.

The mass of the planet is unknown. After 2⁢s, the object has fallen 30m. Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg5⁢kg object near the planet’s surface?
(A) 5 N
(B) 15 N
(C) 37.5 N
(D) 75 N

1 answer:

Umnica [9.8K]2 years ago

8 0

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

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Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

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note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

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and obtain:

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$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

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Plugin this time on our first equation we find:

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2 years ago

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Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

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tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

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<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

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<u>Step 4:</u> calculate the final temperature of the water

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Substitute in the values; we will have,

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W = F dy = F Δy

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F - W = m a

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F = 864 N

The work of this force we will call it W1

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F₂ - W = 0

F₂ = W

F₂ = 80 9.8

F₂ = 784 N

The work of this fura we will call them W2

The total work is

W_total = W₁ + W₂

W_total = (F + F₂) y

W_total = (864 + 784) 5

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W₁ / W₂ = F y / F₂ y

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Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

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