The acceleration of an object would increase if there was an increase in the

Physics

1 answer:

xxMikexx [17]3 years ago

4 0

As per Newton's II law we know that

$F_{net} = ma$

here we know that

$F_{net} = F_{ap} - F_f$

so here we will have

$a = \frac{F_{ap} - F_f}{m}$

so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.

So here correct answer will be

<em>A) force on the object.</em>

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Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a

Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

Vf = √( 2 g X)

Vf = √(2 9.8 1)

Vf = 4.43 m/s

This is the speed with which it reaches the ground

Having the final speed we can find the time

Vf = Vo + g t

t = Vf / g

t = 4.43 / 9.8

t = 0.45 s

This is the time of fall of the body to touch the ground

3 0

3 years ago

A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second

Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$