Answer: Because of Inertia
Explanation: According to Newton first law, that a body in a linear motion will continue to be in motion except an external for act on it
But also, a body in motion will always like to continue in motion and have a reluctance tendency to stop because of inertia.
the crash test dummies' bodies will continue to move forward even though the car stopped because of inertia.
At -40.
-40 gives the same reading for Fahrenheit and Celsius scale.
Answer:

Explanation:
The formula for potential energy is:

where <em>m </em>is the mass, <em>g</em> is the gravitational acceleration, and <em>h</em> is the height.
The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

Substitute the values into the formula.

Multiply the first two numbers.
- 0.4 kg*9.8 m/s²= 3.92 kg*m/s²
- If we convert the units now, the problem will be much easier later on.
- 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

Multiply.
- 3.92 N* 2 m=7.84 N*m
- 1 Newton meter is equal to 1 Joule (this is why we converted the units).
- Our answer is equal to<u> 7.84 Joules.</u>

A) Agreed.
<span>b) Value agreed but units should be W (watts). </span>
<span>c) Here's one method... </span>
<span>15 miles = 24140 m </span>
<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>
<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>
<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>
<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>
<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>
<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>