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Ludmilka [50]
3 years ago
11

Why does it take a large truck longer to stop than a compare car

Physics
1 answer:
mina [271]3 years ago
7 0
Because the truck has more mass
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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
Two point charges exert a 5.50 N force on each other. What will the force become if the distance between them is increased by a
vodomira [7]

Answer:

0.055 N

Explanation:

From coulomb's law,

F α 1/r²

F = k/r²

F₁r₁² = F₂r₂²......................... Equation 1

Where F₁ = Initial force, r₁ = initial distance, F₂ = Final force, r₂ = Final distance.

making F₂ the subject of the equation

F₂ = F₁r₁²/r₂²..................... Equation 2

Given: F₁ = 5.50 N.

Let: r₁ = x m, then r₂= 10x m

Substitute into equation 2

F₂ = 5.5(x²)/(10x)²

F₂ = 5.5x²/100x²

F₂ = 5.5/100

F₂ = 0.055 N.

Hence the force becomes 0.055 N

8 0
3 years ago
Scientific models have two basic types. Please select the best answer from the choices provided T F
zubka84 [21]

Answer:

Scientific models have two basic types. FALSE.

Hoped I helped

3 0
4 years ago
What is the total number of components into which a single force can be reso
anygoal [31]
Two components (vertical and horizontal)

6 0
4 years ago
What is the work done by a 20 newton force applied at an angle of 45.0° to move a box a horizontal distance of 40 meters?
bearhunter [10]
Your answer would be B.

<span>votre réponse serait B</span>
8 0
4 years ago
Read 2 more answers
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