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ankoles [38]
3 years ago
5

Mary applies a force of 70 N to push a box with an acceleration of 0.50 m/s2. When she increases the pushing force to 78 N, the

box's acceleration changes to 0.68 m/s2. There is a constant friction force present between the floor and the box.
(a) What is the mass of the box in kilograms?
(b) What is the coefficient of kinetic friction between the floor and the box?

Physics
2 answers:
qwelly [4]3 years ago
6 0

Answer:

The answers are: a) m = 44.44 kg; b) μ = 0.1097

Explanation:

a)

we have the following data:

F = applied force

f = friction force

F - f = net force = ma

F1 = 70 N

a1 = 0.5 m/s^2

F2 = 78 N

a2 = 0.68 m/s^2

F1 - f = ma1

F2 - f = ma2

Subtracting the second equation from the first equation:

F1 - F2 = m(a1 - a2)

m = (F1 - F2)/(a1 - a2) = (70 - 78)/(0.5 - 0.68) = 44.44 kg

b)

f = μmg

F1 - μmg = ma1

Clearing μ:

μ = (F1 - ma1)/(mg) = [70 - (44.44*0.5)]/[44.44*9.8] = 0.1097

neonofarm [45]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter
kifflom [539]

Answer:

a). 87.5 mA or 87.5 x10^{-3}A

b). 1.78 \frac{m}{s}

Explanation:

d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\  I=87.5 x10^{-3}A

b).

I=n*abs (q)*V_{d}*A

A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2}  \\A=5.309x10^{-6}

V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}

8 0
3 years ago
What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff
sergiy2304 [10]

The net force on the object as described is; 58.84N

Two forces acting on the object are;

  • The <em>applied force and the frictional force.</em>

In essence; the frictional force can be evaluated as;

  • Frictional force; = coefficient × Weight of object.

  • Frictional force = 0.21 × 20 × 9.8.

  • Frictional force = 41.16N

  • The Net force = Applied force - frictional force

  • Net force = 100 - 41.16N

Net Force = 58.84 N.

Read more:

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GalinKa [24]
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Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.
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