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Anna71 [15]
4 years ago
15

The 60-mm-diameter shaft is made of 6061-T6 aluminum having an allowable shear stress of τallow = 80 MPa. Determine the maximum

allowable torque T . Also, find the corresponding angle of twist of disk A relative to disk C .
Physics
1 answer:
grin007 [14]4 years ago
5 0
The maximum allowable torque must correspond to the allowable shear stress for maximization. To solve this, we use the torsion formula:

Max. Allowable Shear Stress = Maximum Torque ÷ Cross-Sectional Area
8 x 10^6 Pa = Maximum Torque ÷ pi*(d/2)²
Maximum Torque = 8 x 10^6 Pa * pi*(0.06/2)² m²
Maximum Torque = 22,619.47 J or
Maximum Torque = 22.62 kJ

As for the second question, I have no reference figure so I am unable to answer it. I hope I was still able to help you, though.
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If a motorcyclist accelerates from rest to 40 m/s in 8 seconds, what is their acceleration?
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Answer:

5 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 40 m/s

t = 8 s

Find: a

a = (v − v₀) / t

a = (40 m/s − 0 m/s) / 8 s

a = 5 m/s²

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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

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Answer:

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Minerals Are Inorganic. ...

Minerals Are Solids. ...

Definite Chemical Composition. ...

Crystalline Structure.

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Explanation:

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Answer:

Momentum

Explanation:

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Momentum is the product of mass and velocity of a body;

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We see that the body with more mass and velocity will have a higher momentum.

Since the two bodies have the same mass, and the speed A is greater, then A will have a greater momentum.

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