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3 years ago

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Cross section of a solid circular rod is subject to a torque of t = 2.7 kn⋅m . if the diameter of the rod is d = 8 cm , what is

the maximum shear stress?

1 answer:

polet [3.4K]3 years ago

3 0

Shear Stress= Force along the tangent/ Area
= 2700/π*r^2
=2700/3.14*16
=54

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Which set of terms best defines what affects kinetic energy and potential energy, respectively? Choose one best answer

bulgar [2K]

Answer:

Velocity, height

Explanation:

6 0

3 years ago

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same am

Tcecarenko [31]

Answer:

Incomplete question,

This is the complete question

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration ~a of the block after it begins to move. Express your answer in terms of some or all of the variables µs, µk, and m, as well as the acceleration due to gravity g.

Explanation:

Let the force that make the object to start moving be F,

Frictional force is opposing the motion, the body has to overcome two frictional forces acting in the opposite direction of the motion.

Also, weight and normal reaction are acting in vertical direction, the weight is acting downward while the reaction is acting upward.

Weight of the object is given as

W=mg

Analyzing the vertical motion i.e y-axis.

ΣF = ma

since the body is not moving upward, the a=0

N-W=0

Then, N=W

So, N=mg

So, from friction law

Fr=µN

For static

Fs=µsN

For kinetic or dynamic

Fk= µkN

Using newton law

Along x-axis

Before the body start moving we can get the Force and since the force is the same use to start the block in motion.

Then,

ΣF = ma

Since at static the body is not moving then, a=0

F-Fs=0

F=Fs

Since, Fs=µsN

F=Fs=µsN

Then, the force to keep the body in motion too is F=µsN

Now analyses when the body is in motion

ΣF = ma

F-Fk=ma

ma=F - Fk

Substituting F=µsN and Fk=µkN

ma=µsN - µkN

ma=N(µs - µk)

Since N=mg

Then, ma=mg(µs - µk)

m cancels out, then

a=g(µs - µk)

Then the acceleration of the body is given as "a=g(µs - µk)"

5 0

3 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

8 0

3 years ago

The first step when doing<br> an<br> investigation is the observe a situation. True or false?

Sunny_sXe [5.5K]

Answer:

True! First step is to make objective observations.

6 0

4 years ago

Read 2 more answers

Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of

zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, $m=40\ kg$

Radius of the barrel is, $R=2.90\ m$

Number of revolutions are, $n =10$

Time taken for 10 revolutions is, $t=29.3\ s$

Therefore, the time period of the child is given as:

$T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s$

Now, angular velocity is related to time period as:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s$

Now, centripetal force acting on the child is given as:

$F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N$

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0

3 years ago