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Ludmilka [50]
4 years ago
6

A block in the shape of a rectangular solid has a cross-sectional area of 5.24 cm2 across its width, a front-to-rear length of 1

5.7 cm, and a resistance of 977 Ω. The block's material contains 4.45 × 1022 conduction electrons/m3. A potential difference of 39.9 V is maintained between its front and rear faces. (a) What is the current in the block
Physics
1 answer:
DerKrebs [107]4 years ago
4 0

Answer:

The current through the block is 40.8mA

Explanation:

For question a we can apply the macroscopic form of Ohm's law:

I=\frac{V}{R}

where:

V= potential difference

R= The resistance of the block

substituting the values on the formula we have:

I=\frac{39.9V}{977 \Omega}\\\\I=40.8mA

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Everyday activities do NOT produce significant muscle growth because:
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When elements bond together or when bonds of compounds are broken and form a new substance

Explanation:

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An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
3 years ago
A steel rod is pulled in tension with a stress that is less than the yield strength. The modulus of elasticity may be calculated
pshichka [43]

Answer:

B. Axial stress divided by axial strain

Explanation:

Elasticity:

It is the tendency of an object to deform along the axis when an opposing force is applied without facing permanent change in shape.

Plasticity:

When an object crosses the elasticity limit, it enters plasticity where the change due to stress is permanent and the object might even break.

Yield strength:

Yield strength is the point of maximum bearable stress that indicates the limit of elasticity.

Our case:

As the stress applied is less than the yield strength, the rod is still in the elasticity state and its modulus can be calculated.

Modulus of Elasticity = Stress along axis/Ratio of change in length to original length

Axial strain is basically the ratio of change in length to original length.

So, Modulus of Elasticity = Axial Stress/ Axial Strain

6 0
4 years ago
Consider two tubes filled with water at the same height, one with fresh water and the other tube with salt water. The pressure i
Olegator [25]

Answer:

B

Explanation:

The correct option for the question is B that is salt water. In salt water, the density of water is higher so the pressure at the end of tube containing salt water will be greater. As according to the hydrostatic law the pressure at a given point will be directly proportional to the distance travelled as well.

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