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3 years ago

A wheel with radius 0.487 m rotates 5.75 times every second. find the period of this motion.

1 answer:

3 0

Calculate the circumference of the circular path of the wheel by the equation,
C = 2πr
where C is the circumference,
r is the radius

Substituting the known values,
C = 2(π)(0.487 m) = 3.06 m

Then, period of the motion will be determined by dividing the number of rotations done,

T = (3.06 m) / (5.75 /s) = 0.53 s

<em>ANSWER: 0.53 s</em>

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DedPeter [7]

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

$L=\dfrac{\mu_o N^2A}{l}$

A is area

$L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH$

(b) The energy stored in the inductor is given by :

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Hence, this is the required solution.

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3 years ago

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Hitman42 [59]

Answer:

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3 years ago

One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.

Zepler [3.9K]

Answer:

The answer is " $\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}$ "

Explanation:

Given:

$\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\$

calculation:

$Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978$

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anyanavicka [17]

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2 years ago

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alexgriva [62]

Answer:

Explanation:

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3 years ago