3 years ago

Pls answer I'm in need lol

Physics

1 answer:

umka2103 [35]3 years ago

4 0

Make into improper fractions. Then make the denominators the same by finding the denominations' lowest common factor.

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the strength of the pulling force on the moon is 1/6 of that on earth whilst on Jupiter it is 2.5 times stronger what is the wei

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Weight is the force exerted on an object by gravity. So, weight of any object on the moon is 1/6 that on Earth.

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The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space

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a parallel circuit has two 8.0 ohm resistors and a power source of 9.0 volts. Of a 12.5 ohm resistor is added to the circuit in

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Read 2 more answers

In an RC-circuit, a resistance of R=1.0 "Giga Ohms" is connected to an air-filled circular-parallel-plate capacitor of diameter

solniwko [45]

Answer:

$\tau = 1\ ms$

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

$C = \epsilon_0 * area / distance$

Where $\epsilon_0$ is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

$Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2$

So the capacitance is:

$C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}$

$C = 10^{-12}\ F = 1\ pF$

The time constant of a rc-circuit is given by:

$\tau = RC$

So we have that:

$\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms$

7 0

3 years ago

The far point of an eye is 180 cm. A corrective lens is to be used to allow this eye to focus clearly on objects a great distanc

sveticcg [70]

Answer:

The focal length of this lens is f = 0.55 D

Explanation:

Given that the far point of an eye ( d )= 180 cm = 1.8 m

We know that focal length of lens is given by

$f = \frac{1}{d}$

Put the value of d in above equation we get

$f = \frac{1}{1.8} \ m^{-1}$

f = 0.55 D

Therefore the focal length of this lens is f = 0.55 D

5 0

3 years ago