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Novay_Z [31]
3 years ago
15

If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

three significant figures and include appropriate units.
Physics
1 answer:
-BARSIC- [3]3 years ago
8 0

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm

Moment of inertia from neutral axis will be given by \frac {bd^{3}}{12}+ ay^{2}

Therefore, moment of inertia will be

\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}

Bending stress at top= \frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa

Bending stress at bottom=\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03 Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

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A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-
joja [24]

Answer:

5.25\cdot 10^{40} kg m^2/s

Explanation:

The angular momentum of the pulsar is given by:

L=m\omega r^2

where

m=2.8\cdot 10^{30} kg is the mass of the pulsar

r = 10.0 km = 1\cdot 10^4 m is the radius

\omega is the angular speed

Given the period of the pulsar, T=33.5\cdot 10^{-3} s, the angular speed is given by

\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s

And so, the angular momentum is

L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s

8 0
3 years ago
What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?​
givi [52]

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

6 0
2 years ago
Which of these is a good example of Newton's First Law of inertia
Eddi Din [679]

Answer: The correct option is A.

Explanation:

Inertia is a state of an object or body to maintain its state. It resists any change in its state.

Newton's first law of inertia: When an objects is in state of motion, it will remain its state of motion or if it is in state of rest then it will remain in rest unless it is acted upon by external motion.

In the given options, a ball sits motionless on the ground is good example of Newton's first law motion. No external force is acting in this case.

In options (B), (C) and (D) , the external force is acting.

Therefore, the correct option is A.

8 0
3 years ago
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Anastaziya [24]

Answer:

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it has potential energy relative to its position on the floor, However, if the object were to fall it would hit the floor with a KE equal to the PE that it had sitting on the shelf.

Sounds are caused by compressional waves in  the air - when a piano key is struck or a TV is turned on, then compressional waves are produced in the surrounding air due to a disturbance. The human ear recognizes the disturbed air as due to the object that created the disturbance.

7 0
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