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3 years ago

A circuit is supplied with 30 volts and the total circuit resistance is 3 kΩ. What's the current flowing through the circuit?

Physics

1 answer:

Daniel [21]3 years ago

4 0

Answer: The current flowing through the circuit is 0.01A (or 10 mA)

Explanation:

Use Ohm's Law:

$I=\frac{U}{R}$

Given the values of U=30V and R=3000Ohm:

$I=\frac{30 V}{3000 \Omega}=0.01A=10mA$

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A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r

kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = $K.E=\frac{1}{2}I\omega^2$

where,

I is the moment of inertia = mr²

on substituting the values, we get

$K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2$

K.E = 0.0075 J

3 0

3 years ago

Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?

vagabundo [1.1K]

<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

Then, as they accelerate toward Earth (regardless of their mass), it curves beneath them and never comes close.

That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

8 0

3 years ago

Read 2 more answers

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

Tpy6a [65]

Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

7 0

4 years ago

If the distance between two masses is tripled, the gravitational force between changes by a factor of

maw [93]

A. 1/9

Explanation:

The gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1 and m2 are the two masses

r is the distance between the two masses

From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

$\frac{1}{r^2}=\frac{1}{3^2}=\frac{1}{9}$