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Hunter-Best [27]
3 years ago
15

What happens in terms of energy when a moving car hits a parked car, causing the parked car to move?

Physics
2 answers:
Helen [10]3 years ago
8 0

the answer to this question is c

meriva3 years ago
5 0
This is an example Newton's Third Law. All the kinectic energy from the moving car transferred the potential energy of the parked car. This potential is not much since the brakes are on (hopefully) and it's not in a non-moving position.
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7. Using Analogies Explain how the behavior
klemol [59]

The relative motion of gaseous particles increases with increase in the temperature of the gas molecules just like the motion of popcorn in a popper increases when heat is applied to the popper.

<h3>What is kinetic theory of gas?</h3>

The kinetic theory of gases or matter states that matter consists of tiny particles which are constant motion, colliding with one another and with walls of the containing vessels.

Just like a popcorn in a popcorn popper pops when heat is applied to the popper, gases contained in a cylinder increases their speed when they acquire more kinetic energy as the temperature of the cylinder increases.

Thus, the motion of gas particles depends on the temperature of the containing vessel so also does the random motion of popcorn depends on the temperature of the popper.

Learn more about kinetic theory of gases here: brainly.com/question/11067389

#SPJ1

7 0
8 months ago
Not a question, but I need help on the last two questions I posted, preferably no links, please...
Andrei [34K]

Answer:

ok...

Explanation:

7 0
3 years ago
Read 2 more answers
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
2 years ago
Based on Schaffer's arguments in "The Value of a Sherpa Life," choose the statement
Tju [1.3M]

Answer:

Sherpas do work that is much more meaningful than the work other climbers do.

Explanation:

3 0
3 years ago
Part A Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.7 m whose uniform depth is 1.8 m . Express y
Sphinxa [80]

Answer:

17.66 kPa

Explanation:

The volume of water in the swimming pool is the product of its dimensions

V = 30 * 8.7 * 1.8 = 469.8 cubic meters

Let water density \rho = 1000 kg/m^3, and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool

W = mg = \rho V g = 1000 * 469.8 * 9.81 = 4608738 N

The area of the bottom

A = 30 * 8.7 = 261 square meters

Therefore the pressure is its force over unit area

P = F/A = 4608738  / 261 = 17658 N/m^2 or 17.66 kPa

7 0
3 years ago
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