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3 years ago

8

The moon’s rotational period and revolution period are equal. What is the result of this? Enter your answer in the space provide

d.

1 answer:

beks73 [17]3 years ago

5 0

Answer The Moon has synchronous rotation: it's rotation period is the same as its period of revolution

Explanation:

The Moon has synchronous rotation: it's rotation period is the same as its period of revolution

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A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?

pochemuha

10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv

7 0

4 years ago

Figure 1 shows a wave movement during one second. What is the frequency of the wave

Andreyy89

This is 2 hertz. You can mark out 2 full wavelengths in the second of time.

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4 years ago

Read 2 more answers

The tent meteorite, found in 1897 near cape York, on the west coast of Greenland, is the largest meteorite exhibited by any muse

sweet-ann [11.9K]

Answer:
3.0883 x 10^10mg

Explanation:
1 kilogram = 1000 000 milligrams
So, 30 883 x 1000 000 = 30 883 000 000mg

6 0

3 years ago

When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc

Nesterboy [21]

Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

Explanation:

We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

w² = mg d / I

In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow

d = L / 2

The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

I = ¼ m r2 + ⅓ m L2

I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

ρ = m / V

m = ρ V

the volume of a cylinder is

V = π r² L

m = ρ π r² L

let's substitute

w² = m g (L / 2) / m (¼ r² + ⅓ L²)

w² = g L / (½ r² + 2/3 L²)

L >> r

w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

4 0

4 years ago

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

Tasya [4]

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0

3 years ago