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diamong [38]
3 years ago
13

The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C

lockwise as the negative direction.

Physics
1 answer:
Mashcka [7]3 years ago
8 0

Answer:

Force A=-−2,697.75 N

Force B=13, 488.75 N

Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

25 mg-20Fb=0

25*1100g=20Fb

Fb=25*1100g/20=1375g

Taking g as 9.81 then Fb=1375*9.81=13,488.75 N

The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g

-275*9.81=−2,697.75. Therefore, force A pulls downwards

Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle

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Answer:

The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

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Time t = 0.14 s

We need to calculate the flux

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\phi=NAB

Put the value into the formula

\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5

\phi=0.003180

We need to calculate the emf

Using formula of emf

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{0.003180}{0.14}

\epsilon =-0.000227\ V

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Negative sign shows the direction of current.

Hence, The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

8 0
4 years ago
What is the role of magnetism in the study of x-ray imaging?
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8 0
3 years ago
Two objects of mass 6 kg and 3 kg respectively move at a constant speed of 5m/s in a straight line. When an equal external force
Ivanshal [37]
A is the correct answer
8 0
3 years ago
What are the factors affect the Electric forces between two charges and What is the relationship between each factor and the Ele
Cloud [144]

Explanation:

If the two charges are point charges - i.e., they don't have a size - the force between these charges depends on the

• Magnitude if each charge, q1 and q2

• Sign of each charge (+ or -)

• Distance between the charges, r

This is essentially Coulomb’s Law:

FE = (kq1q2)/r2

For collections of charges, you need to find the electric field E, and then use this fields to find a force on a small test charge q in the field. The test charge is always small to help you map the electric field, but not disturb it.

3 0
3 years ago
An experiment on the earth's magnetic field is being carried out 1.00 m from an electric cable. What is the maximum allowable cu
Whitepunk [10]

Answer:

The allowed current in the cable is 1.15 A.

Explanation:

Given that,

Distance = 1.00 m

Suppose the magnetic field is 2.3\times10^{-5}\ T and  if the experiment is to be accurate to 1.0 %

We need to calculate the current

Using formula of magnetic field

B=\dfrac{\mu_{0}I}{2\pi r}

I=\dfrac{B\times2\pi r}{\mu_{0}}

Put the value into the formula

I=\dfrac{2.3\times10^{-5}\times2\times\pi\times1.00}{4\pi\times10^{-7}}

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If the experiment is to be accurate to 1.0%

Then,

We need to calculate the allowed current in the cable

I'=(1.0%)\times I

I'=\dfrac{1.0}{100}\times115

I'=1.15\ A

Hence, The allowed current in the cable is 1.15 A.

4 0
3 years ago
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