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hodyreva [135]
3 years ago
11

using mass and distance, identify and compare the sun's and moon's contribution to the formation of tides on earth

Physics
1 answer:
vekshin13 years ago
7 0

Answer:

Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.

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By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and
puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

m_1=43.7 kg

m_2=12.1 kg

g=9.8 m/s^2

a=\frac{1}{2}(9.8)=4.9 m/s^2

We have to find the mass of the pulley.

According to question

T_2-m_2 g=m_2 a

T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N

T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N

Moment of inertia of pulley=I=\frac{1}{2}Mr^2

(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)

Where \alpha=\frac{a}{r}

(177.87-214.13)=-\frac{1}{2}(4.9)M

-36.26=-\frac{1}{2}(4.9)M

M=\frac{36.26\times 2}{4.9}=14.8 kg

Hence, the mass of the pulley=14.8 kg

6 0
3 years ago
The table describes the properties of four compounds.
irga5000 [103]

Answer:

compounds C and D

Explanation:

I believe this is correct

8 0
3 years ago
Read 2 more answers
Anytime an object is moving, it has what kind of energy
irina1246 [14]
Any object that is moving has Kinetic energy
7 0
3 years ago
when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its c
andrey2020 [161]

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

          ΔA = 2α A ΔT

the area is

         A = π r² = π D² / 4

we substitute

         ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

    α = 1.7 10⁻⁵ ºC⁻¹

we calculate

          ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

          ΔA = 6.49 10⁻⁷ m²

by definition

           ΔA = A_f- A₀

           A_f = ΔA + A₀

           A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

           A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

           A_f = 2,551 10⁻⁴ m²

the area is

           A_f = π D_f² / 4

           A_f = \sqrt{4  A_f /\pi }

           D_f = \sqrt{4 \ 2.551 10^{-4} /\pi }

           D_f = 1.80229 10⁻² m

the change in diameter is

           ΔD = D_f - D₀

           ΔD = (1.80229 - 1.8) 10⁻² m

           ΔD = 0.00229 10⁻² m

           ΔD = 2.29 10⁻⁵ m

5 0
3 years ago
How does water get up to the atmosphere, and how does it get back down to earth surface
IrinaK [193]

Answer:

Water gets up to the Earth's atmosphere by evaporating from a body of water, which is then they become water vapor. It returns back to the surface by returning back to its water state and falling back down (as rain). The water vapor turns into clouds (clouds are really just water droplets), and when it cannot hold anymore waters, it disperses all the water (by raining).

6 0
3 years ago
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