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V125BC [204]
1 year ago
8

suppose you have a 69.0-kg wooden crate resting on a wood floor. what maximum force can you exert horizontally on the crate with

out moving it?
Physics
1 answer:
marusya05 [52]1 year ago
7 0

You've got a 69.0-kg wooden crate on a wooden floor. The box can withstand a force of up to 338N in a horizontal direction without being moved. Following this, the wooden creates moving stats.

In order to calculate the friction coefficient, divide the force pushing two objects together by the force acting between them. friction coefficient might be 0 or one. They can be split into two categories: friction coefficient that is static. Kinetic friction coefficient (also known as sliding coefficient of friction).

the acceleration brought on by the gravitational pull of large masses generally, gravitational , often known as the acceleration brought on by the Earth's gravitational pull and centrifugal force,

F= friction coefficient *M*g

F= 0.5*69*9.8

F=338N

Learn more about gravitational here

brainly.com/question/3009841

#SPJ4

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The nucleus of most atoms is composed of which of the following sub-atomic particles?
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3 years ago
A 50 g mass is freely hanging from a horizontal meter stick at a distance of 99 cm from the pivot. Calculate the weight force W
Neko [114]

Answer:

W = 0.49 N

τ = 0.4851 Nm

Force

Explanation:

The weight force can be found as:

W = mg

W = (0.05 kg)(9.8 m/s²)

<u>W = 0.49 N</u>

The torque about the pivot can be found as:

τ = W*d

where,

τ = torque

d = distance between weight and pivot = 99 cm = 0.99 m

Therefore,

τ = (0.49 N)(0.99 m)

<u>τ = 0.4851 Nm</u>

The pivot exerts a  <u>FORCE </u>on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.

6 0
3 years ago
Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t
Orlov [11]

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

\tau = Fdsin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

\theta is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so \theta=90^{\circ}, sin \theta=1

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

4 0
3 years ago
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