All categories

3 years ago

Jill is pushing a box across the floor. Which represents the upward force perpendicular to the floor?

Physics

1 answer:

AfilCa [17]3 years ago

7 0

I think it fb I’m not really sure

You might be interested in

Why does a solid change to liquid when heat is added?

Komok [63]

Answer:

c.the spacing between particles increases

3 0

3 years ago

Read 2 more answers

What do we call the study of the science of movement

nalin [4]

We call it: Biomechanics

3 0

4 years ago

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

nataly862011 [7]

Answer with Explanation:

Let rest mass $m_0$ at point P at distance x from center of the planet, along a line connecting the centers of planet and the moon.

Mass of moon=m

Distance between the center of moon and center of planet=D

Mass of planet=M

We are given that net force on an object will be zero

a.We have to derive an expression for x in terms of m, M and D.

We know that gravitational force= $\frac{GmM}{r^2}$

Distance of P from moon=D-x

$F_m$ =Force applied on rest mass due to m

$F_m$ =Force on rest mass due to mas M

$F_M=F_m$ because net force is equal to 0.

$F_m=F_M$

$\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}$

$\frac{m}{(D-x)^2}=\frac{M}{x^2}$

$\frac{x^2}{(D-x)^2}=\frac{M}{m}$

$\frac{x}{D-x}=\sqrt{\frac{M}{m}}$

Let $R=\sqrt{\frac{M}{m}}$

Then, $\frac{x}{D-x}=R$

$x=DR-xR$

$x+xR=DR$

$x(1+R)=DR$

$x=\frac{DR}{1+R}$

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x= $\frac{2}{3}D$

Net force is zero

$F_m=F_M$

$\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}$

$\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}$

$\frac{M}{m}=4$

Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

8 0

4 years ago

Definition: The energy transferred by a force to a moving object.

Marysya12 [62]

Answer:

Force applied

Explanation:

An object will remain at its state of rest unless a non zero for act on it

6 0

3 years ago

1. A 65 kg student, starting from rest, slides down an 16.2 m high water slide. On the way down, friction does 5700 J of work on

GuDViN [60]

Answer:

11.94

Explanation:

Remark

Find the Potential Energy at the top.

Givens

m = 65 kg

h = 16.2 m

g = 9.81

PE = 65 * 9.81 * 16.2

PE = 10329.93

The tricky part is what do you do about Friction?

Formula

PE = Friction + KE

Solution

PE = 10329.93 Joules

Friction = 5700 Joules

Find the KE

10329.93 = 5700 + KE

KE = 10329.93 - 5700

KE = 4629.93

Find V from the KE formula

KE = 4629.93

m = 65

KE = 1/2 m v^2

KE = 1/2 65 v^2

4629.93 = 1/2 65 v^2

v^2 = 142.46

v = √142.46

v = 11.94

4 0

3 years ago