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kow [346]
3 years ago
11

An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed

northbound on the same track at 1 m/s. The faster train bumps into the slower one, and they combine into one train. What is the velocity of the two-car train?
Physics
1 answer:
Tema [17]3 years ago
9 0

Let the mass of 2500 kg car be m_1 and it's velocity be v_1 and the mass of 1500 kg car be m_2 and it's velocity be v_2 .

After the bumping the mass be M and it's velocity be V.

     By law of conservation of momentum we have

                   m_1v_1+m_2v_2 = MV

                    2500 * 5 + 1500 * 1=4000 * V

                    V = 14000/4000 = 7/2 = 3.5 m/s

So the velocity of the two-car train = 3.5 m/s

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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

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

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

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

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
The S-strain Pneumococcus bacteria had a smooth surface, because ____________________.
drek231 [11]
The S strain Pneumococcus bacteria had a smooth surface because IT IS SURROUNDED BY A CARBOHYDRATE CAPSULE CALLED THE S STRAIN. The other form, the R strain has a rough surface and no capsule. It is only the S strain that exhibits virulence. 
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The resistance of the body varies from approximately 500 kΩ (when it is very dry) to about 1.00kΩ (when it is wet). The maximum
Nady [450]

Answer:

you absolute buffoon Use Ohms' Law:   V = RI

V = (1x10^3)(5x10^-3) = 5 volts

Yes, this is in the range of normal household voltages.  

Explanation:

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3 years ago
If the force of gravity changes what will happen and what would cause it to change?
FromTheMoon [43]
We will fall to pluto.
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3 years ago
A ball is dropped from a building taking 3sec to fall to the ground. Calculate:
GenaCL600 [577]

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

  • Initial Velocity (Vo) = 0 m/s
  • Gravity (g) = 9.8 m/s²
  • Time (t) = 3 s
  • Final Velocity (Vf) = ?
  • Height (h) = ?

==================================================================

Final Velocity

Use formula:

  • Vf = g * t

Replace:

  • Vf = 9.8 m/s² * 3s

Multiply:

  • Vf = 29.4 m/s

==================================================================

Height

Use formula:

  • \boxed{h=\frac{g*(t)^{2}}{2}}

Replace:

  • \boxed{h=\frac{9.8\frac{m}{s^{2}}*(3s)^{2}}{2}}

Multiply time squared:

  • \boxed{h=\frac{9.8\frac{m}{s^{2}}*9s^{2}}{2}}

Simplify the s², and multiply in the numerator:

  • \boxed{h=\frac{88.2m}{2}}

It divides:

  • \boxed{h=44.1\ m}

What is the velocity when falling to the ground?

The final velocity is <u>29.4 meters per seconds.</u>

How high is the building?

The height of the building is <u>44.1 meters.</u>

3 0
3 years ago
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