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8_murik_8 [283]
1 year ago
7

What is the moons gravitational field ?

Physics
1 answer:
ollegr [7]1 year ago
8 0

Answer:

approximately 1.625 m/s , about 16.6% that on Earth's surface or 0.166 ɡ. Over the entire surface, the variation in gravitational acceleration is about 0.0253 m/s.

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If a cylindrical space station 275 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) mus
sergij07 [2.7K]

The centripetal acceleration is responsible for the artificial gravity because the acceleration of an object moving in constant circular motion causing from net external force is called centripetal acceleration. It defines to the center or seeking the center.

Given the following:

Cylindrical space station diameter           = 275 meters; 137.5 meters for the radius

Standard gravity                                       = 9.80665 m/s²

 

Using the formula:

w² x r =g

w² = g / r

w² = 9.80665 m/s² / 137.5 m

w² = 9.80665 m/s² / 137.5 m

w² = 0.0713 s²

Then take the roots

w = 0.267 this is radians per second / 2 x (3.1416 which is the pi)

w = 0.0424 rps convert to rpm

w = 0.0424 r/s (1minute / 60 seconds)

w = 7.08 x 10⁻⁴ revolutions per minute

4 0
4 years ago
Benny can’t get the metal lid off a new glass jar of jam. His mother tells him to run hot water carefully over the lid. He does
ELEN [110]

Answer:

The cap expanded

Explanation:

As you run hot water (Or any substance) onto something it expands, With coldwater it shrinks. Its like clothes, If a shirt is to small wash it in hot water. If it is to big run it in cold water.

6 0
3 years ago
What are political obstacles keeping wind turbine from being used?
erastova [34]

Explanation:

Turbines might cause noise and aesthetic pollution.

Although wind power plants have relatively little impact on the environment compared to conventional power plants, concern exists over the noise produced by the turbine blades and visual impacts to the landscape.

The inflexibility, variability, and relative unpredictability of wind power as a means for electricity production, are the most obvious barriers to an easy integration and widespread application of wind power." Thus, the uncertainty of the wind requires a system that is always available to replace all the electrical output created by the wind turbine system. In other words, it is too expensive to have wind turbines lying around that don't do anything.

7 0
3 years ago
As outlaws escape in their getaway ear, which goes 3/4 c, the police officer fires a bullet from a pursuit ear, which only goes
Mariulka [41]

Answer:

a) Bullet will hit

b) Bullets will not hit

Explanation:

Given:

The velocity of the bullet, u = \frac{1}{3}c in the rest frame of the bullet pursuit car

The velocity of the original frame of reference, v = -\frac{1}{2}c with respect to the pursuit car.

Now, according to the Galileo

the velocity of the bullet in the original frame of reference (u') will be

u' = u - v

on substituting the values we get

u' = \frac{1}{3}c-(-\frac{1}{2}c)

or

u' = \frac{1}{3}c+\frac{1}{2}c

or

u' = \frac{5}{6}c

since this velocity ( \frac{5}{6}c) is greater than the ( \frac{3}{4}c)

hence,

<u>the bullet will hit</u>

Now, according to the Einstein theory

the velocity of the bullet in the original frame of reference (u') will be

u'=\frac{u-v}{1-\frac{uv}{c^2}}

on substituting the values we get

u'=\frac{\frac{1}{3}c-\frac{1}{2}c}{1-\frac{\frac{1}{3}c\times \frac{1}{2}c}{c^2}}

or

u'=\frac{\frac{5}{6}c}{1-\frac{1}{6}}

or

u'=\frac{5}{7}c

since,

u'=\frac{5}{7}c is less than  ( \frac{3}{4}c), this means that the bullet will not hit

7 0
4 years ago
A 105 kg football player runs at 8.5 m/s and plows into the back of an 85 kg referee running at 3.5 m/s on the field causing the
Vedmedyk [2.9K]

Answer:

680 Kg.m/s

Explanation:

Mass of player; m_p = 105 kg.

Speed of player before Collision; v_pi = 8.5 m/s

Mass of referee; m_r = 85 kg

Speed of referee before collision; v_ri = 3.5 m/s

Speed of referee after collision; v_rf = 6 m/s

From conservation of momentum,

Initial momentum = final momentum

Thus;

(m_p × v_pi) + (m_r × v_ri) = (m_p × v_pf) + (m_r × v_rf)

Where (m_p × v_pf) is the momentum of the player after collision.

Thus, Plugging in the relevant values, we have;

(105 × 8.5) + (85 × 3.5) = (m_p × v_pf) + (85 × 6)

(m_p × v_pf) = (105 × 8.5) + (85 × 3.5) - (85 × 6)

(m_p × v_pf) = 680 Kg.m/s

3 0
3 years ago
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