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3 years ago

Can someone help me?

Physics

1 answer:

Leviafan [203]3 years ago

4 0

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

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Which statement best compares momentum and kinetic energy?

Katyanochek1 [597]

<span>If your options are:

A.Both momentum and kinetic energy are vector quantities.

B.Momentum is a vector quantity and kinetic energy is a scalar quantity.

C.Kinetic energy is a vector quantity and momentum is a scalar quantity.

D.Both momentum and kinetic energy are scalar quantities.
</span>
The answer on the question given is letter B.<span>Momentum is a vector quantity and kinetic energy is a scalar quantity.</span>

4 0

3 years ago

Explain the difference between the synodic period of a planet and the sidereal period. why is the synodic period often much less

Ilia_Sergeevich [38]

-Synodic period is the period of celestial bodies observed on the moving planet(mostly earth)
Sideral period is the period comparing to the fixed stars without motion of the earth involved.

(I will explain the second question with an example, so it's easier to understand)
-For Sideral month for example of the moon it cactually complete one revolution in around 27.3 days.
However, since the earth moves, for us it took some more time to see the moon the same as before (fullmoon to fullmoon) again. That make synodic month of the moon to be around 29.5 days.

4 0

4 years ago

A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo

CaHeK987 [17]

Answer: $\omega =93.51 rad/s$

Explanation:

Given

mass of Flywheel $m_1=1440 kg$

mass of bus $m_b=10200 kg$

radius of Flywheel $r=0.63 m$

final speed of bus $v=21 m/s$

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let $\omega$ be the angular velocity of Flywheel

$0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}$

$I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2$

$0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}$

$\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}$

$\omega =21\times 4.45=93.51 rad/s$

8 0

3 years ago

A 65.8-kg person throws a 0.0413 kg snowball forward with a ground speed of 32.5 m/s. A second person, with a mass of 58.7 kg, c

guapka [62]

Answer:

$v_{1} = 2.490\,\frac{m}{s}$

$v_{2} = 0.023\,\frac{m}{s}$

Explanation:

The statement is described physically by means of the Principle of Momentum Conservation. Let assume that first person moves in the positive direction:

First Person

$(65.8\,kg)\cdot (2.51\,\frac{m}{s}) = (65.8\,kg)\cdot v_{1} + (0.0413\,kg)\cdot (32.5\,\frac{m}{s} )$

Second Person

$(0.0413\,kg)\cdot (32.5\,\frac{m}{s})+(58.7\,kg)\cdot (0\,\frac{m}{s})=(0.0413\,kg+58.7\,kg)\cdot v_{2}$

The final velocities of the two people after the snowball is exchanged is:

$v_{1} = 2.490\,\frac{m}{s}$

$v_{2} = 0.023\,\frac{m}{s}$

6 0

3 years ago

Which statement correctly describes the gravitational potential energy of the pendulum based on this diagram?

MrRa [10]

I think it's 'C' but I won't know for sure until you let me see the diagram.

8 0

4 years ago

Can someone help me?​

Can someone help me?