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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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Sequence the steps used to extract metal from the earths core
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4 0
3 years ago
A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou
zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}

E=6.88\times10^{7}\ N/C

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}

E=5.43\times10^{6}\ N/C

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0
3 years ago
The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the
Ronch [10]

Answer: 1

Explanation:

Given

Tension is the string T=44\ N

mass of object m=4\ kg

Tension is greater than the weight of the object i.e. elevator is moving upward

we can write

\Rightarrow T-mg=ma\\\Rightarrow T=m(g+a)\\\Rightarrow 44=4(10+a)\\\Rightarrow 11=10+a\\\Rightarrow a=1\ m/s^2

8 0
3 years ago
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