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nydimaria [60]
3 years ago
6

ndicate whether the statement is true or false. Whenever we move, we alter the rate at which we move into the future. A. True B.

False
Physics
1 answer:
Ksju [112]3 years ago
4 0
Whenever we move, we alter the rate at which we move into the future. This statement is true. 
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Compared to yellow light, orange light has
Mkey [24]
<span>So we want to know what does orange light has when we compare it to yellow light. So when we take a look at the electromagnetic spectrum we can see that orange color has longer wavelength than yellow color. So the correct answer is A. alonger wavelength.</span>
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3 years ago
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Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and
Lelu [443]

Answer:

The maximum shear stress in shaft AB, T_{ABmax} is 15 MPa

The maximum shear stress in shaft CD,  T_{CDmax} is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J = \pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}

Therefore, we have

J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}

Where:

D_{AB} = Diameter of shaft AB = 30 mm = 0.03 m

r_{AB} = Radius of shaft AB = 15 mm = 0.015 m

∴ J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2} = 7.95 × 10⁻⁸ m⁴

and

J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}

Where:

D_{CD} = Diameter of shaft CD = 36 mm = 0.036 m

r_{CD} = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2} = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= 1.58 \times \frac{2\pi }{360} rad = 2.76 × 10⁻² rad.

That is \phi_r = 2.76 × 10⁻² rad.

However  \phi_r =  \phi_{C/D} -  \phi_{B/A}  

Where:

\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G} and

\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}

T_{AB} and T_{CD}= Torque on shaft AB and CD respectively

T_{AB}  = Required

T_{CD}= 500 N·m

L_{AB} and L_{CD} = Length of shafts AB an CD respectively

L_{AB}  = 600 mm = 0.6 m

L_{CD} = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

\phi_r =  \phi_{C/D} -  \phi_{B/A}  =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

2.76 × 10⁻² rad =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

=\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}

Therefore;

T_{AB} =  79.54 N.m

Where T = T_{AB} + T_{CD} =

Therefore T_{CD total } = 500 - 79.54 = 420.46 N·m

τ_{max} = \frac{T\times R}{J}

\tau_{ABmax} = \frac{T_{AB}\times R_{AB}}{J_{AB}} =  \frac{79.54\times 0.015}{7.95\times 10^{-8}} = 15 MPa

\tau_{CDmax} = \frac{T_{CD}\times R_{CD}}{J_{CD}} = \frac{420.46\times 0.018}{1.65\times 10^{-7}} = 45.9 MPa

7 0
3 years ago
Imagine that you pushed a box, applying a force of 60 newtons, over a distance of 4 meters. how much would you have done? 15 jou
Ray Of Light [21]
The answer is 240 joules. Good luck!
8 0
3 years ago
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Ayudaaa no se como resolverlo con procedimiento :(
Lubov Fominskaja [6]
This is not Spanish bro
6 0
3 years ago
What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co
MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0
3 years ago
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