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3 years ago

How fast does water flow from a hole at the bottom of a very wide, 3.2 m deep storage tank filled with water

Physics

1 answer:

Nezavi [6.7K]3 years ago

5 0

Answer:

the velocity of the water flow is 7.92 m/s

Explanation:

The computation of the velocity of the water flow is as follows

Here we use the Bernouli equation

As we know that

$V_1 = \sqrt{2g(h_2 - h_1)}\\\\=\sqrt{2g(\Delta h)} \\\\= \sqrt{2\times9.8\times3.2} \\\\= \sqrt{62.72}$

= 7.92 m/s

Hence, the velocity of the water flow is 7.92 m/s

We simply applied the above formula so that the correct value could come

And, the same is to be considered

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An AC generator has an output rms voltage of 100.0 V at a frequency of 42.0 Hz. If the generator is connected across a 45.0-mH i

Rus_ich [418]

Answer:

(a) 11.8692‬ ohm

(b) 12.447 A

Explanation:

a) inductive reactance Z = L Ω

= L x 2π x F

= 45.0 x 10⁻³ x 2(3.14) x 42

= 11.8692‬ ohm

b) rms current

= 100 / 8.034

= 12.447 A

c) maximum current in the circuit

= I eff x rac2

= 12.447 x 1.414

= 17.6 A

4 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is: