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Vitek1552 [10]
2 years ago
12

A gravitational force of 2.54 x 10^-10 is exerted by two objects that are 0.25 m apart. If the first object has a mass of 0.35 k

g, what is the mass of the second object?
Physics
1 answer:
Murrr4er [49]2 years ago
3 0

Answer :

what is the answer options?

Explanation:

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The strength of the force of friction depends on which two factors?
Stolb23 [73]

Options A and D are correct. The strength of the force of friction depends on the objects' sizes and weights and the heat generated by the friction and the types of surfaces involved.

<h3 /><h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration in the different components and balancing the equation gets. Components in the x-direction.

The strength of the force of friction depends on the two factors, as;

A. The objects' sizes and weights.

D. The heat generated by the friction and the types of surfaces involved.

Hence, options A and D are correct.

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3 0
2 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
Explain how machines can be useful if the output is always less than the input work
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Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.

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<h2>Virtual image</h2>

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4 0
3 years ago
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Gravity is affected by...
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Mass and distance
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