Does anyone know the answer to this page and the next one that it has?

Cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, ini

kakasveta [241]

The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s

<h3>Conservation of Linear Momentum</h3>

Given Data

Mass of cart one M1 = 150kg

Initial Velocity U1 = 8m/s

Final VelocityV1 = 5 m/s

Mass of cart two M2 = 150kg

Velocity U2 = 6m/s

Applying the principle of conservation of linear momentum we have

M1U1+M2U2 = M1V1+ M2V2

a. what is the speed of cart b after collision

substituting our given data we have

150*8+ 150*6 = 150*5+150*V2

1200 + 900 = 1200+ 150V2

2100 - 1200 = 150V2

900 = 150V2

Divide both sides by 150

V2 = 900/150

V2 = 6m/s

b. what is the total momentum of the system before and after collision

Total Momentum in the system is

Total momentum = Momentum before Impact+ Momentum after Impact

Total momentum = M1U1+M2U2 + M1V1+ M2V2

Total momentum = 1200 + 900 + 1200+ 900

Total momentum = 4200 kg m/s

Learn more about Conservation of Linear Momentum here:

brainly.com/question/7538238

6 0

2 years ago

The unit for measuring electric power is the <br> A. ampere.<br> B. volt.<br> C. ohm.<br> D. watt.

Drupady [299]

The correct answer is D: Watt. This unit was named after James Watt, and is used to express the equivalent of one joule per second in energy. In experiments and on the packaging for electrical products such as light-bulbs, the measurement will usually be written in its abbreviated format: W.
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6 0

3 years ago

Can someone help me pls

Dmitry_Shevchenko [17]

First question: 800J
Second question: 20.4m

4 0

3 years ago

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

$\rho = 12580.7 kg/m^3$

8 0

2 years ago

A pendulum is hanging from a tower. The pendulum almost touches the ground and has a period of 18.0 s. What is the height of the

nadya68 [22]

We apply the following equation

T = 2π * sqrt (L/g)

Where g is the gravity = 9.8 m/s^2

L is the longitude of the pendulum (Height of the tower)

T is the period. (T = 18s)

We find L.............> (T /2π)^2 = L/g

L = g*(T /2π)^2...........> L = 80.428 meters

5 0

3 years ago