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3 years ago

8

50POINTS! What is the escape velocity for lunar module? Lunar module mass 15,200 kg radius of moon 1.74x106m, mass of moon 7.34x

1022kg

2 answers:

Airida [17]3 years ago

7 0

Answer:

2.73 km/s

Explanation:

The escape velocity of an object in the gravitational field of the moon is (on the surface of the planet)

$v=\sqrt{\frac{2GM}{r}}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=7.34\cdot 10^{22} kg$ is the mass of the Moon

$r=1.74\cdot 10^6 m$ is the radius of the Moon

As we can see, the escape velocity does not depend on the mass of the lunar module.

Substituting the numbers into the formula, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(7.34\cdot 10^{22}kg)}{(1.74\cdot 10^6 m)}}=2732 m/s=2.73 km/s$

mash [69]3 years ago

6 0

Answer:

2.73 km/s

Explanation:

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A 10 g bullet accelerates to 400 m/s after being fired from a gun with a mass of 2.0 kg.

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Answer:

See the answers below.

Explanation:

Momentum is defined as the product of mass by velocity, and can be calculated by means of the following expression.

$P=m*v$

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P = Momentum [kg*m/s]

m = mass = 10 [g] = 0.01 [kg]

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Identify the following physical quantities as scalars or vectors.

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4 years ago

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A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, $\lambda_1=8\times 10^{-6}\ C$

linear charge density of the other charge distribution, $\lambda_2=-6\times 10^{-6}\ C$

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.r.\epsilon_0}$

where:

$\lambda=$ linear charge density

r = radial distance from the center of wire

$\epsilon_0=$ permittivity of free space

Therefore,

$E_1=E_2$

$\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}$

$\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}$

$\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}$

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When a car is traveling 120 km/h how many times more kinetic energy would it have than if it were

LekaFEV [45]

Answer: 9

Explanation:

Kinetic energy is the energy possessed by a moving object. It is measured in joules, and depends on the mass (m) of the object and the speed (v) by which it moves i.e K.E = 1/2 x mass x velocity^2

So, car is traveling 120 km/h

KE = 1/2x m x (120 km/h)^2

KE = 0.5 x m x 14400

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So, car is traveling 40 km/h

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