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2 years ago

In the equation for the gravitational force between two objects, which quantity must be squared?

Physics

1 answer:

bezimeni [28]2 years ago

8 0

Answer:

Explanation:

The quantity that must be squared in the equation of gravitational force is distance d.

According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.

Therefore, the quantity to be squared is d

The formula is given as:

Fg = $\frac{G m_{1} m_{2} }{d^{2} }$

So d is the quantity that must be squared

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A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

3 years ago

Work-Energy Theorem: A 4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects en

allsm [11]

The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:

FΔx = 0.5mv²

F = force, Δx = distance traveled, m = mass, v = speed

Isolate Δx:

Δx = 0.5mv²/F

Calculate Δx for each object.

Object 1: m = 4.0kg, v = 2.0m/s

Δx = 0.5(4.0)(2.0)²/F = 8/F

Object 2: m = 1.0kg, v = 4.0m/s

Δx = 0.5(1.0)(4.0)²/F = 8/F

The two objects travel the same distance before stopping.

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2 years ago

Solving for acceleration,mass and net force

Dvinal [7]

The formulas are-

a=f/m
f=ma
m=f/a

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2 years ago

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What is the best summary for this cartoon?

Paha777 [63]

Answer:

C: World War I resulted from a chain reaction involving many nations.

Explanation:

Just did the questions and got it right

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2 years ago

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How does a gas do work?

Triss [41]

Your answer would be C, "By conserving energy"! A gas can do work through something called the First Law of Thermodynamics. Heat that energy used by the gas during this "work" cannot be created or destroyed however it can be transferred and converted. This energy can only be stored during these processes, not removed. An example of this would be gas in a cylindrical object such as a piston. When the gas is heated, it expands and moves the piston upwards. When it expands, it stores that energy it gains through the heat given.

I hope this helps! (:

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3 years ago

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