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3 years ago

The rectangular coils in a 345-turn generator are 12 cm by 12 cm. Part A

Physics

1 answer:

Likurg_2 [28]3 years ago

4 0

Answer:

118.166 volt

Explanation:

We have given number of turns N =345

Sides of the rectangular coils is 12 cm =0.12 m

So area A =0.12×0.12=0.0144 $m^2$

Magnetic field B =0.45 T

Angular speed =505 rpm

Speed in rad/sec $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 505}{60}=52.8566rad/sec$

The emf is given by $E=NAB\omega SIN\omega t$

For maximum emf sinwt =1

So $E=345\times 0.0144\times 0.45\times 52.8566=118.166volt$

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21. Calculate the acceleration of the bus from point D to E. Show your work.

Marat540 [252]

21) Acceleration from D to E: $1 m/s^2$

22) The acceleration of the bus from D to E is $1 m/s^2$

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

$a=\frac{5-0}{5}=1 m/s^2$

22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

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4 0

3 years ago

The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve

solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d = 0.107 mm = 0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 / 4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0

3 years ago

4. During which three months is the difference the same between average high temperature and average low temperature?

kkurt [141]

Answer:

1 Inch and 20 F

Explanation:

6 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant