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2 years ago

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A step-down transformer has 2500 turns on its primary and 5.0 x 10' tums on its secondary. If the potential difference across th

e primary is 4850 V, what is
the potential difference across the secondary?

1 answer:

Brilliant_brown [7]2 years ago

4 0

Answer:

I dont know sorry

Explanation:

hehe

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A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>

Vadim26 [7]

Answer:

$\huge\boxed{\sf f=0.55 \ Hz}$

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }$

<u>Solution:</u>

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}$

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2 years ago

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

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How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'

Serhud [2]

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We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.

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2 years ago