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lilavasa [31]
3 years ago
13

If the moon is rising at midnight, the phase of the moon must be

Physics
1 answer:
Nadya [2.5K]3 years ago
7 0
The phase is called 3rd quarter.

Hope this helps:)
You might be interested in
A cart for hauling ore out of a gold mine has a mass of 413 kg, including its load. the cart runs along a straight stretch of tr
Andrei [34K]
Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle  @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N;  the component of the force from the donkey in the direction of motion isF2 = F1 [email protected]= 397.2964498768165 N
Fy = N - mg [email protected] = 0N = mg [email protected] = 4037.964151113007 NFx = F2 - mg [email protected] - f = mahere f = u N=65.8188156631420141

F2 - mg [email protected] - f = maa = F2 - mg [email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J
8 0
3 years ago
4. The time it takes an object to rotate once is ​
pshichka [43]

Answer:

Explanation:

23.9344696 hours

3 0
3 years ago
Based on the type of medium, what can you assume about the speed of sound in glass?
adelina 88 [10]

Answer:

"It will be more than the speed of sound waves in air at 20*C and water at 20*C."

Explanation:

Speed of sound in a medium depends upon the density and elasticity of the medium.

If the elasticity of a medium is greater and the density of that medium is lower, sound will travel faster. Although density is also a factor but the major factor is Elasticity.

Hence, sound travel faster in solids than in liquids and even slower in gases due to elasticity difference.

                        Speed of sound = V = \sqrt{\frac{Elasticity}{Density} }

That is why the speed of sound in glass will be more than the speed of sound waves in air at 20*C and water at 20*C.

5 0
3 years ago
5. A cable is attached 32.0 m from the base of a flagpole that is about to
soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, F_v = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

F_v + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈  4513.093·d  

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0
2 years ago
Two people push on a boulder to try to move it. The mass of the boulder is 825 kg. One person pushes north with a force of 64 N.
stira [4]

Answer:

0.09 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

From the question, expression for acceleration is given as

F' = ma  

Using Pythagoras Theory,

√(F₁²+F₂²) = ma................... Equation 1

Where F₁ = Force of the First person on the boulder, F₂ = Force of the Second person on the boulder, F' = resultant force acting on the boulder, m = mass of the boulder, a = acceleration of the boulder.

make a the subject of the equation

a = √(F₁²+F₂²) /m................ Equation 2

Given: m = 825 kg, F₁ = 64 N, F₂ = 38 N,

Substitute into equation 2

a = [√(64²+38²)]/825

a = {√(5540)}/825

a = 74.43/825

a = 0.09 m/s²

4 0
3 years ago
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