Ask question

Ask question

All categories

stealth61 [152]

3 years ago

9

a 77 N net force is applied to a box which slides horizontally across a floor for 6.7 m. What amount of work is done on the box

by the net force ?

1 answer:

ArbitrLikvidat [17]3 years ago

7 0

W=fd
W=77*6.7
W=515.9 J

You might be interested in

A kilowatt-hour is a unit of

AnnZ [28]

Answer:

option 4 is the correct answer

6 0

3 years ago

Read 2 more answers

Why do you think compound machines are sometimes called complex machines

Maksim231197 [3]

Answer:

i think maybe because it has more than 6 or 5 parts.

3 0

4 years ago

Consider the following statements. A. Heat flows from an object at higher temperature to an object at lower temperature; B. Heat

Monica [59]

Answer:

A. Heat flows from an object at higher temperature to an object at lower temperature

Explanation:

The option A obeys the 2nd law of thermodynamics. The heat will flow from the object at higher temperature to the object at Lower temperature till they reach an equilibrial state.

Heat doesn’t necessarily flow from an object with higher thermal energy to an object with lower thermal energy because an object has a higher thermal energy when it’s mass is more than the other. This makes B wrong.

C is wrong because heat moves from an object with higher temperature to objects with Lower temperature regardless of the state of matter.

8 0

3 years ago

A force of 500 N acting on an object of 2 kg displaces it by 3 m. Find the work done by the force and the object's final speed a

Salsk061 [2.6K]

Answer:

buscalo en gugul nby buenas nochis

7 0

3 years ago

Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 12

ikadub [295]

Answer:

The unknown mass of the bullet is $m1=3.751x10^{-3} kg$

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass <u><em>m</em></u> , it results in change in momentum of the body and is given by:

$F=\frac{P}{dt}$

Where:

P

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

$m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\$

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

$m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\$ $m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\$

$-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg$

3 0

3 years ago