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solniwko [45]
3 years ago
14

Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi

ng with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Physics
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

<u>W = 12.96 J</u>

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m mass
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What factors contribute to global winds identify areas where winds are weak
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Two tuning forks A and B when sounded together produce 5 beats per second. The tuning fork A has frequency 480Hz. The fork B is
BartSMP [9]
<h2>The frequency of tuning fork B is 475 Hz</h2>

Explanation:

The number of beats produced is equal to the frequency difference between the two tuning forks .

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Thus the frequency of B can be = 485 or 475 Hz

When B is filed , its frequency increases and it start producing 2 beats with A .

Thus it frequency must be 475 Hz , by filing , it increases to 478 Hz .

By which it gives  480 - 478 = 2 beats per second .

This cannot happen with frequency 485 , because by filing this difference increases in place of decreasing .

5 0
3 years ago
A bowling ball weighing 71.2 N (16.0 lb) is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and release
Elina [12.6K]

Answer:

a) For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2

b) For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

T= mg +ma

In order to find the mass we know that W=mg and solving for m we got:

m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg

And replacing into the tension we got:

T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N

Explanation:

For this case we have the following data given:

W= 71.2 N represent the weigth for the object

R=3.8 m the length of the rope or the radius

v= 4.2 m/s represent the velocity of the bowling ball

Part a

For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2

Part b

For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

T= mg +ma

In order to find the mass we know that W=mg and solving for m we got:

m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg

And replacing into the tension we got:

T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N

7 0
2 years ago
A very long straight wire carries a 12 A current eastward and a second very long straight wire carries a 14 A current westward.
s344n2d4d5 [400]

Answer: 5.12x10∧-4N

Explanation:

Force = I B L

L = 6.4m

Let Current (I) I₁ = I₂= 14A

Distance of the wire = 42cm = 0.42m

BUT

B = μ₀I / 2πr

=(2X10∧-7 X 12) / 0.42

       B =5.714×10∧-6T

Force = I B L

Force = 14x [5.714×10-6]×6.4

Force = 5.12x10∧-4N

7 0
3 years ago
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