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bonufazy [111]
3 years ago
15

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

er. what is the magnitude of the normal force between the piston and cylinder? (b) what is the magnitude of the force would she have to exert if the steel parts were oiled?
Physics
1 answer:
Vilka [71]3 years ago
8 0
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
since we know that F=300 N and \mu=0.3, we can find N, the magnitude of the normal force:
N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
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2 years ago
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Place gamma rays, infrared, microwaves, radio waves, ultraviolet, visible light, and x-rays in order from largest wavelength to
Brums [2.3K]

Answer:

Going by EM SPECTRUM WE HAVE

radio waves, microwaves, infrared, VISIBLE LIGHT, ultraviolet, X-rays, GAMMA RAYS

Explanation:

BECAUSE

V= WAVELENGTH/ FREQUENCY

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7 0
3 years ago
An electric kettle. The input energy is 10 Joules. The useful output energy is heat 9
mina [271]

Answer:

Efficiency = 90 %

Wasted energy = 10 %

Explanation:

Since we have the input energy and useful output energy of the electric kettle, the only thing we are required to calculate here is its efficiency. This is gotten from

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E = 9/10 × 100 = 90 %

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3 0
3 years ago
A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft

Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft

Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

d = BD + CD = 70.4 + 93.4\\d = 163.8 ft

5 0
3 years ago
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