All categories

3 years ago

When ____________________ equals the force of gravity on a falling object, the object reaches terminal velocity.

Physics

1 answer:

Lilit [14]3 years ago

5 0

The correct answer to the question is : Air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand the terminal velocity.

An object is said to be moving with terminal velocity if the net force acting on the object is zero i.e the body is non accelerating.

As per the question, the object is falling under gravity. The force of gravity acts in vertical downward direction.

Except gravity, the object is also under air resistance. It is the force which opposes the motion of the object, and acts in vertical upward direction.

If the air resistance becomes equal to force of gravity, then the net force acting on the object is zero . It is so because two forces act in opposite direction.

In case of liquids, the resistance force is called as viscosity.

Hence, the correct answer to the question is air resistance.

You might be interested in

Who would have the most potential energy?

snow_lady [41]

Answer:

I'm pretty sure it's A.

4 0

3 years ago

Read 2 more answers

A 4 kg billiard ball moving on a horizontal surface has a speed of 16 m/s when it strikes a horizontal coiled spring is brought

ruslelena [56]

Answer:

spring constant of the spring is 1820.44 N/m

Explanation:

given data

ball mass = 4 kg

speed = 16 m/s

distance = 0.75 m

to find out

spring constant of the spring

solution

we know that kinetic energy of ball = energy store in spring as compression

so we can express it as

0.5 × m × v² = 0.5 × k × x² ....................1

so put here value we get spring constant k

m × v² = k × x²

4 × 16² = k × 0.75²

solve it we get

k = 1820.44 N/m

so spring constant of the spring is 1820.44 N/m

6 0

4 years ago

Read 2 more answers

Two massive objects are fixed in position. A third object is placed directly between the first two at the position at which the

ivanzaharov [21]

Answer:

Explanation:

Be M1 and M2 the two massive objects and m the third one

The gravitational force over the third object at the beginning is zero (assuming that the position of the third object is x=0)

$F_{m}=-G\frac{mM_{1}}{x_{1}^{2}}-G\frac{mM_{2}}{x_{2}^{2}}=0$

when the third object is displaced we have that the new position is x=Δx. Hence

$F_{m}=-G\frac{mM_{1}}{(-x_{1}+\Delta x)^{2}}-G\frac{mM_{2}}{(x_{2}-\Delta x)^{2}}$

I attached a scheme of the system

I hope this is useful for you

Regards

5 0

3 years ago

How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00

laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

$Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J$

4 0

4 years ago

Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface

nekit [7.7K]

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

$sin\theta_c=\frac{n_2}{n_1}$

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

$n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30$

hope this helps!!

6 0

3 years ago