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tankabanditka [31]
3 years ago
6

The two spheres pictured above have equal densities and are subject only to their mutual gravitational attraction. Which of the

following quantities must have the same magnitude for both spheres? (A) Acceleration (B) Velocity (C) Kinetic energy (D) Displacement from the center of mass (E) Gravitational force........ I understand the answer is E but dont know how to explain it
Physics
1 answer:
kiruha [24]3 years ago
8 0

Answer:

Gravitational force

Explanation:

If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.

It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.  

Hence, the correct option is (E) "Gravitational force".                        

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Light striking a mirror at a 50° angle will be reflected at an angle _____.
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If a liquid has a density of 1.67 g/cm^3, what is the volume of 45g of the liquid?
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Answer:

V = 26.95 cm³

Explanation:

Density is given by the formula :

ρ = m÷V

Density = mass ÷ Volume

Given both density and mass we rearrange, substitute and solve for Volume :

Rearranging the equation to make Volume the subject :

ρ = m÷V

ρV = m

V = m÷ ρ

Now substitute :

V = 45 ÷ 1.67

V = 26.9461077844

Take 2 decimal places as the density is 2 decimal places :

V = 26.95

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Hope this helped and have a good day

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3 years ago
6. A warehouse employee is pushing a 15.0 kg desk across a floor at a constant speed of 0.50 m/s. How much work must the employe
MariettaO [177]

Answer:

7.5 J

Explanation:

To answer the question given above, we need to determine the energy that will bring about the speed of 1 m/s. This can be obtained as follow:

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Energy (E) =?

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E = 7.5 J

Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.

3 0
2 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, w
solniwko [45]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

                                   = 0.025×< -2, 0, -7 >

                                   = < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

                                    = 0.025×< 4, 4, -3 >

                                    = < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact  =  < -0.05,  0,  0.175 > + < 0.1, 0.1, -0.075>

                                                      = < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together  after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity =  v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

          = < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

7 0
3 years ago
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