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Alexxx [7]
3 years ago
14

Find the gravitational force between the

Physics
1 answer:
love history [14]3 years ago
7 0

Answer:

The process is given in the pic.

I have taken the average masses so u substitute the values and solve hope it will help :)❤

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((PLEASE HELP))
Degger [83]

Answer:

Angle of refraction

Explanation:

The incident ray is the ray before it reaches the surface.

The refracted ray is the ray after it reaches the surface.

n₁ is called the index of incidence.

n₂ is called the index of refraction.

θ₁ is called the angle of incidence.

θ₂ is called the angle of refraction.

They are related by Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

5 0
3 years ago
Read 2 more answers
You are riding a bicycle home from school. You average 5 meters per second and you are headed south.
olchik [2.2K]

Answer:

(OD) Velocity

Explanation:

Here, the rider is moving with a steady speed (5 m/s) towards south. In this example, we have magnitude as well as direction. Since velocity is a vector quantity, thus we can determine the velocity of the rider.

5 0
3 years ago
1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the
zheka24 [161]

Answer:

a)   F = 4.9 10⁴ N,  b)   F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

           P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

          ∑F = 0

          F-W = 0

          F = W = mg

          F = 5000 9.8

          F = 4.9 10⁴ N

the area of ​​the pisto is

          A = pi r²

          A = pi d² / 4

          A = pi 1 ^ 2/4

          A = 0.7854 m²

pressure is

          P = 4.9 104 / 0.7854

          P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

          \frac{F_1}{A_1} = \frac{F_2}{A_2}

where subscript 1 is for the small piston and subscript 2 is for the large piston

          F₁ = \frac{A_1}{A_2} \ F_2

the force applied must be equal to the weight of the truck

          F₁ = ( \frac{d_1}{d_2} )^2\  m g

          F₁ = (0.05 / 1) ² 5000 9.8

          F₁ = 122.5 N

7 0
3 years ago
If you're driving towards the sun late in the afternoon, you can reduce the glare from the road by wearing sunglasses that permi
zepelin [54]
The correct option is B.
Sunglasses are designed to block polarized light. If you are driving toward the sun,  most of the sun's glare will be coming to you horizontally, thus, the best sunglasses to wear is the one, that will block the horizontal glare and allow only the vertical glare to pass through. 
3 0
3 years ago
Read 2 more answers
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

T=41.85\ N  

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0
3 years ago
Read 2 more answers
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