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bezimeni [28]
3 years ago
8

a horizontal force of 100N is required to push a crate across a factory floor at a constant speed. What is the net force acting

on the crate/ What is the force of friction acting on the crate?
Physics
1 answer:
Bas_tet [7]3 years ago
8 0

If the crate is moving along the floor in the same direction with a constant speed, it is in dynamic equilibrium. Equilibrium means there is no net force acting on the crate.

Since there is no net force on the crate, there must be a friction force on the crate equal in magnitude and opposite in direction to the applied horizontal force. Therefore the force of friction acting on the crate is 100N.

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Branches of sicence​
Sonja [21]

Answer:

physical science

earth science and life science

5 0
3 years ago
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A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +46 N·m is ap
Rom4ik [11]

Answer:

Explanation:

Let Torque due to friction be

F  

Net torque

= 46 - F

Angular impulse = change in angular momentum

=(  46 - F ) x 17  = I X 580

When external torque is removed , only friction creates torque reducing its speed to zero in 120 s so

Angular impulse = change in angular momentum

F  x 120 = I X 580

(  46 - F ) x 17 = F  x 120

137 F = 46 x 17

F = 5.7 Nm

b )

Putting this value in first equation

5.7 x 120 = I x 580

I = 1.18 kg m²

8 0
3 years ago
During an automobile crash test, the average force exerted by a solid wall on a 1,700 kg car that hits the wall is measured to b
Molodets [167]

Answer:

14 m/s

Explanation:

Step One:

Given data:

mass of automobile m= 1700kg

Force F= 170,000 N

time t= 0.14s

v= ??

Required

The velocity of the car

Step two:

From the expression given below, we can find the velocity

Ft= mv

make v subject of the formula

v= Ft/m

v= 170000*0.14/1700

v=23800/1700

v=14 m/s

3 0
3 years ago
A constant voltage is applied across a circuit element. If the resistance of the element is doubled, what is the effect on the p
miv72 [106K]

Answer:

The power dissipated is reduced by a factor of 2

Explanation:

The power dissipated by a resistor is given by:

P=I^2 R

where

I is the current

R is the resistance

by using Ohm's law, I=\frac{V}{R}, we can rewrite the previous equation in terms of the voltage applied across the resistor (V):

P=(\frac{V}{R})^2R=\frac{V^2}{R}

In this problem, the resistance of the element is doubled, while the voltage is kept constant. So we have R'=2R while V remains the same; substituting into the formula, we have:

P'=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{P}{2}

so, the power dissipated is reduced by a factor 2.

5 0
3 years ago
The diagram shows the electric field around two charged objects.
anyanavicka [17]

Here we have been given two charged body W and X.

We are asked to determine the nature of charge.

Before coming into conclusion, first we have to understand the electric field lines.

The electric field lines are pictorial representation of imaginary lines which are drawn to denote the electric field in a graphical  way.

The electric field line starts from a positive charge and ends at a negative charge.It means for positive charges the electric field lines are outward and for negative charges the filed lines will be inward.

In the given diagram,the filed lines for W is towards W itself.The same is also in case of X.

Hence both the charges must be negative in nature.

Hence the correct answer to the question will be B i.e W negative and X: NEGATIVE.

8 0
3 years ago
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