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s344n2d4d5 [400]
3 years ago
10

Please help me asdfghjklqwertyuioikjhgfrfghjuhygfdfghjmhgtfhjk

Physics
1 answer:
sertanlavr [38]3 years ago
5 0

Answer:

c

Explanation:

the moon moves around earth

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Which of the following relationships is correct?
masha68 [24]

Answer:

a. Relative humidity increases: a decrease in temperature without a change in the amount of water vapor in the atmosphere.

Explanation:

Colder air requires less humidity to saturate compared to warmer air. Therefore, the relative humidity depends on the air temperature. If the temperature drops and the water vapor content remains the same, relative humidity increases. If the temperature increases and the water vapor content remains the same, relative humidity decreases. Additionally, a greater amount of water vapor in the air increases the relative humudity.

3 0
3 years ago
A student has two 150kΩ resistors and two 200ߤF capacitors from which to create an RC circuit. Determine each of the possible ci
FrozenT [24]

Answer:

The time constants 30 seconds and the can form an RC Charging circuit.

Explanation:

Time constant, τ, is measured in terms of τ = R x C, in seconds, where R is the value of the resistor in ohms and C is the value of the capacitor in Farads

If R = 150 kΩ = 150000Ω

and C = 200 μF = 0.0002 F

τ = 150,000 × 0.0002

<u>= 30 seconds.</u>

A capacitor, (C)  connected in series with a resistor, (R) forms a RC Charging Circuit connected across a DC battery supply (Vs) via a mechanical switch, with transient time (time required to charge capacitor) = 5τ

3 0
3 years ago
A 20,000 kg jet plane is accelerating at 2 m/s2. How much force does it put forth?
Alex17521 [72]

Answer:

40,000N of force

Explanation:

The formula for force is F=ma. If we plug in the values, F=(20,000kg)(2m/s^2).

8 0
3 years ago
Read 2 more answers
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.12 m whose potential is 230
slamgirl [31]

Answer:

(a) q=3.07 nC

(b) σ=17 nC/m²

Explanation:

Given data

Radius r=0.12m

Potential V=230 V

To find

(a) Charge q

(b) Charge density σ

Solution

For Part (a)

As we know that potential is:

V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}

Substitute the given values

q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC

For Part (b)

The charge density is given by:

σ=q/(4πr²)

Substitute the given values and value of q to find charge density

So

=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2

σ=17 nC/m²

3 0
3 years ago
The surface temperature of our sun is about 5800 K, and the peak of its intensity curve is in the middle of the visible spectrum
amm1812

Explanation:

i expected to use Stefan's law of heat exchange but the value you gave aren't conclusive.

I should say that the temperature of the star should be close to that of the sun because of the similarity in the intensity curves

3 0
2 years ago
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