Ask question

Ask question

All categories

2 years ago

14

Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe

ed of 0.850 m/s and air track cart B has a mass of 300 g and moves toward the left with a speed of 1.12 m/s. What is the total momentum of the system?
A) 0.719 kg•m/s toward the right
B) 0.719 kg•m/s toward the left
C) 0.750 kg•m/s toward the right
D) 0.047 kg•m/s toward the right
E) 0.750 kg•m/s toward the left

1 answer:

Blizzard [7]2 years ago

6 0

Answer:

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

Explanation:

Hi there!

The total momentum of the system is given by the sum of the momentum vectors of each cart. The momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Then, the momentum of the system will be the momentum of cart A plus the momentum of cart B (let´s consider the right as the positive direction):

mA · vA + mB · Vb

0.450 kg · 0.850 m/s + 0.300 kg · (- 1.12 m/s) = 0.047 kg · m/s

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

You might be interested in

What type of glass absorbs light and does not allow light to pass through it?

sesenic [268]

opaque glass does not allow light to pass through it.

8 0

3 years ago

What is four possible energy sources for a circuit

VashaNatasha [74]

<span>electric, solar, wind, and geothermal.</span>

8 0

3 years ago

Read 2 more answers

A child pushes a toy car down a hill. The child has a mass of 20 kg. The car has a mass of 1.6 kg and a speed of 7.4 m/s2. When

Andrew [12]

Answer:

The answer is 73.8 J

Explanation:

5 0

3 years ago

Several short trips taken from a cold start can use ....... as much fuel as a longer multi-purpose trip covering the same distan

Lerok [7]

Several short trips taken from a cold start can use ...twice... as much fuel as a longer multi-purpose trip covering the same distance when the engine is warm.

In cold weather, properly designed gasoline aids in engine starting, while in hot weather, it helps prevent vapor lock. In order to meet the requirements of a modern engine, the fuel must have the volatility for which the engine's fuel system was built and an antiknock quality strong enough to prevent knock during routine operation.

During the intake phase, the air and fuel are combined before being introduced into the cylinder. The spark ignites the fuel-air mixture after the piston compresses it, resulting in combustion. During the power stroke, the piston is propelled by the expansion of the combustion gases.

To learn more about engine and fuel please visit -
brainly.com/question/5181209
#SPJ4

4 0

1 year ago

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0

3 years ago

Read 2 more answers