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Gnesinka [82]
3 years ago
8

A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo

r and box? Determine it to three significant figures even through that's quite unrealistic. How much work is done in overcoming friction between the object and floor if the box slides 8m along horizontally on the floor?
Physics
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

Coefficient of friction is 0.068.

Work done is 320~J.

Explanation:

Given:

Mass of the box (m): 60 kg

Force needed (F): 40 N

The formula to calculate the coefficient of friction between the floor and the box is given by

F=\mu mg...................(1)

Here, \mu is the coefficient of friction and g is the acceleration due to gravity.

Substitute 40 N for F, 60 kg for m and 9.80 m/s² for g into equation (1) and solve to calculate the value of the coefficient of friction.

40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068

The formula to calculate the work done in overcoming the friction is given by

W=Fd..........................(2)

Here, W is the work done and d is the distance travelled.

Substitute  40 N for F and 8 m for d into equation (2) to calculate the work done.

W=40~N\times8~m\\~~~~= 320~J

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sesenic [268]

Answer:

The power dissipated in a resistor is 117.54 watts.

Explanation:

Given that,

Peak voltage of the Ac generator, V = 230 V

Frequency, f = 210 Hz

Resistance, R = 225 ohms

We need to find the power dissipated in a resistor. The power generated is given by :

P=\dfrac{V^2_{rms}}{R}

V_{rms}=\dfrac{V}{\sqrt{2} }\\\\V_{rms}=\dfrac{230}{\sqrt{2} } \\\\V_{rms}=162.63\ V

So,

P=\dfrac{162.63^2}{225}\\\\P=117.54\ W

So, the power dissipated in a resistor is 117.54 watts. Hence, this is the required solution.

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3 years ago
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1.Radio waves from using your TV.

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4.x ray waves used at the doctors office

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While working on her science fair project Venus connected a battery to a circuit that contained a light bulb. Venus decided to c
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4 years ago
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3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

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- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
4 years ago
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