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3 years ago

Why are large astronomical bodies round

2 answers:

8 0

I would have to go with gravity, as each object in this universe has some amount of gravity, even humans and creatures and plants.

when a rock is thrown into space, it will have to travels light years away and pull in thousands of other asteroids into yet sometimes this doesn't work and sometimes nothing will happen depending on the speed of how they are going.

collision of these rocks are also another cause to the round shape, remember these rocks could be thousands of millions of years old. and this would be easy for them to shape because of the amount of asteroids in space.

this shape is also by the gravity on it in of from it. though this rare and it will be called a meteorite.

-Dominant- [34]3 years ago

7 0

<span>Matter of all types have gravity, which causes it to attract to each other. The most efficient way for all this matter to congregate is the sphere. As they consolidate, they form the shape. As they compress, temperatures in the center of the mass start to go up and if it hits the proper point, it can ignite and become a star.</span>

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Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20

timama [110]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

3 0

3 years ago

Read 2 more answers

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

3 years ago

How many times louder is the intensity of sound at a rock concert in comparison with that of a whisper, if the two intensity lev

Phoenix [80]

Answer:

5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper

Explanation:

Given that at a rock concert;

the intensity of sound is 110 dB compared to a whisper of 3 dB

Now; how many times louder will that of the whisper be compared to the sound of the rock concert

Using the formula for calculating decibels (dB);

For 110 dB

dB = 10log₁₀(W)

110 dB = $10^{(110/10)}$

110 dB = 10¹¹

For 3dB

dB = 10log₁₀(W)

3 dB = $10^{(3/10)}$

3 dB = 2

Now:

(110/3)dB = 10¹¹/2

(110/3)dB = 5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper

7 0

2 years ago

What causes the difference in the angle of the sun on the Earth's surface throughout the year?

Novosadov [1.4K]

Answer:

The axis is tilted and points to the North Star no matter where Earth is in its orbit. Because of this, the distribution of the Sun's rays changes. ... It also means that the angle at which sunlight strikes different parts of Earth's surface changes through the year.

Explanation:

Pls sub to bdoggaming if this helped

5 0

2 years ago

Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affec

lara31 [8.8K]

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

$F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1$

Therefore, the force between the charges are not affected.

7 0

3 years ago