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3 years ago

7

A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove

rcome to keep the object moving at a constant velocity?

1 answer:

jonny [76]3 years ago

4 0

We must remember that the total net force equation at constant velocity is:

<span>F – Ff = 0</span>

of

F - µN = 0

Using Newton's 2nd Law of Motion:<span>

F = m a

<span>Where,

F = net force acting on the body
m = mass of the body
a = acceleration of the body

Since the cart is moving at a constant velocity, then acceleration is zero, hence the working equation simplifies to

F = net Force = 0

Therefore,

F - µN = 0

where

µ = coefficient of friction = 0.20
N = normal force acting on the cart = 12 N

Therefore,

F - 0.20(12) = 0

<span>F = 2.4 N </span></span></span>

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Light waves from the sun can be converted to electricity through __________.

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Answer:

light waves can be converted to electricity through <em>a solar cell</em>

Explanation:

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3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

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Hence, the ground speed of the plane is 96.68 km/h.

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3 years ago

A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

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$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

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When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

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By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

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Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

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I'm not too sure what your asking but here are two answers that may help.
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Fiesta28 [93]

Answer:

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But :) we are taking infinitesimal small time also

As you know if denominator is small fraction is large So fraction always give large value

So it's not O ( this makes confuse to most of students)

So, thanks

Good question

Keep thinking like this :)

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4 years ago

Read 2 more answers