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3 years ago

A 10kg box putting pushed with a force of 24 Newtons and a fraction of 12N.

Physics

2 answers:

Goshia [24]3 years ago

5 0

Hi! I’ll help you in just a momentwhen I’m done with my walk!

Olin [163]3 years ago

3 0

Answer:

The two dogs sitting here are already poor and ignorant

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What explains the information needed to calculate speed and velocity?

Ede4ka [16]

To calculate speed, you need to know the total distance covered
and the time it took to cover the distance.

To calculate velocity, you need the starting location, the ending location,
and the time it took to go from start to finish.

3 0

3 years ago

In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es

tigry1 [53]

Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

4 0

3 years ago

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

$\omega_f = 0.4\ rad/s$

Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

$\omega_f = 0.4\ rad/s$

8 0

3 years ago

sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o

crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

$F_{net} = F_1 - F_2$