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3 years ago

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The accompanying map shows the route traveled by a school bus.What is the magnitude of te dispacment odte school bus fo start to

te end of its trip?

1 answer:

krok68 [10]3 years ago

4 0

Answer:1,800

Explanation:

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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

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce

Aleksandr [31]

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

$a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]$

where L is the length of one car.

The same equation can be written considering the first 7 cars:

$7L = ut+\frac{1}{2}at'^2$

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

$a=0.08L$

Substituting into the equation, we can find t':

$7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s$

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

A man in a strength competition pulls an 18-wheel truck 3.10 m in 20.5 s. There is a cable that is attached to his body that exe

larisa [96]

Answer:

114.32195122 but Round your answer to three significant figures.) is 114

Explanation:

Just took the test

4 0

3 years ago

Fill in the blank with the appropriate numbers for both electrons and bonds(considering that single bonds are counted as one, do

tekilochka [14]

Answer:

1.Fluorine is having 7 number of electrons and 1 makes bond.

Electronic configuration - 1S² 2S² 2P⁵

1 electron need to get in stable states 2P⁶.

2.Oxygen is having 6 balance electron and 2 makes bonds.

Electronic configuration - 1S² 2S² 2P⁴

2 electron need to get in stable states 2P⁶.

3.Nitrogen is having 5 balance electron and 3 makes bonds.

Electronic configuration - 1S² 2S² 2P³

3 electron need to get in stable states 2P⁶.

4. Carbon having 4 balance electron and 4 makes bonds.

Electronic configuration - 1S² 2S² 2P²

4 electron need to get in stable states 2P⁶.

3 0

3 years ago

Please help!!!!!

adoni [48]

Gpe is basses on the force equation...
GPE=m*g*h=1.5kg*9.8m/s^2*8m=117.6 N*m

8 0

3 years ago

Read 2 more answers