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bonufazy [111]
2 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]2 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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Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u
Aleksandr [31]
Let
upthrust = T
weight = W = mg
Air resistance = F

When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)

Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)

Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
3 0
2 years ago
Which two pieces of data indicate that Uranus resides in the outer region of the solar system
LuckyWell [14K]

Answer:

Our solar system has total eight planets out of which four are inner planets and four are outer planets. The four outer planets are Jupiter, Saturn, Uranus and Neptune. The common characteristics of outer planets is that they are gaseous planets. They are larger on size than the inner rocky planets and are faraway from Sun. They have larger period of revolution around the Sun.

Uranus is a gaseous planet and lies far from Sun and hence has large period of revolution. It takes 84 Earth years to revolve around Sun. This data indicates that Uranus resides in the outer region of the Solar System.

4 0
2 years ago
a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv
Lina20 [59]

Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

So, impulse = 42 N-m and change in momentum =42 N-m.

7 0
2 years ago
A swimmer does 3,560 J of work in 55 s. What is the swimmer’s power output? Round your answer to two significant figures. The po
Natasha2012 [34]
The value of the swimmer's power output is calculated by dividing the work done by the time it took for the work to be completed. From the given in this item,
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Rounding off to two significant figures will give us 65 W. 
6 0
2 years ago
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