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3 years ago

Can anybody help me out???!?!?

Physics

2 answers:

OlgaM077 [116]3 years ago

7 0

Its the C, the sun and the moon, the other planets are too far away to have any gravitational impacts, the moon impacts the heaviest.

Travka [436]3 years ago

6 0

Only the moon is the correct answer.

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What is one factor that helps keep the planets and other solar system objects in orbit around the sun?

pshichka [43]

The answer is D<span>.the gravitational force between the sun and each object in the solar system</span>

7 0

3 years ago

Read 2 more answers

there are two types of waves: electromagnetic and mechanical. can either type travel regardless of the presence of a medium?

Dmitry_Shevchenko [17]

Yes, electromagnetic can travel without medium.

Mechanical waves and electromagnetic waves are two important ways that energy is transported in the world around us.

Waves in water and sound waves in air are two examples of mechanical waves.

Mechanical waves are caused by a disturbance or vibration in matter, whether solid, gas, liquid, or plasma.

Matter that waves are traveling through is called a medium.

These mechanical waves travel through a medium by causing the molecules to bump into each other, like falling dominoes transferring energy from one to the next.

Sound waves cannot travel in the vacuum of space because there is no medium to transmit these mechanical waves.

On the other hand electromagnetic waves don't require medium for its propagation.

An easy example would be light which is an EM wave reaches earth even though space has no medium.

Learn more about different types of waves here:

brainly.com/question/13364787

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3 0

1 year ago

An uncharged series RC circuit is to be connected across a battery. For each of the following changes, determine whether the tim

slavikrds [6]

a) Increase

b) Unchanged

c) Increase

Explanation:

The charge on a capacitor charging in a RC circuit connected to a battery follows the exponential equation:

$Q(t)=Q_0 (1-e^{-\frac{t}{RC}})$

where

$Q_0 = CV$ is the final charge stored in the capacitor, where C is the capacitance and V is the voltage of the battery

t is the time

R is the resistance of the circuit

The capacitor reaches 90% of its final charge when

$Q(t)=0.90Q_0$

Substituting and re-arranging the equation, we find:

$0.90Q_0 = Q_0(1-e^{-\frac{t}{RC}})\\0.90=1-e^{-\frac{t}{RC}}\\e^{-\frac{t}{RC}}=0.10\\-\frac{t}{RC}=ln(0.10)\\t=-RCln(0.10)=2.30RC$

We see that if we double the RC constant, then $(RC)'=2(RC)$

So the time taken will double as well:

$t'=2.30(RC)'=2.30(2RC)=2(2.30RC)=2t$

So, the answer is "increase"

In this second part, the battery voltage is doubled.

According to the equation written in part a),

$Q_0 =CV$

this means also that the final charge stored on the capacitor will also double.

However, the equation that gives us the time needed for the capacitor to reach 90% of its full charge is

$t=2.30 RC$

We see that this equation does not depend at all on the voltage of the battery.

Therefore, if the battery voltage is doubled, the final charge on the capacitor will double as well, but the time needed for the capacitor to reach 90% of its charge will not change.

So the correct answer is

"unchanged"

In this case, a second resistor is added in series with the original resistor of the circuit.

We know that for two resistors in series, the total resistance of the circuit is given by the sum of the individual resistances:

$R=R_1+R_2$

Since each resistance is a positive value, this means that as we add new resistors, the total resistance of the circuit increases.

Therefore in this problem, if we add a resistor in series to the original circuit, this means that the total resistance of the circuit will increase.

The time taken for the capacitor to reach 90% of its final charge is still

$t=2.30 RC$

As we can see, this time is directly proportional to the resistance of the circuit, R: therefore, if we add a resistor in series, the resistance of the circuit will increase, and therefore this time will increase as well.

So the correct answer is

"increase"

8 0

2 years ago

What are ways to improve the design of this experiment? Check all that apply

Serhud [2]

Answer:

B,D,E

Explanation:

I got you

B. Experiment with a wider range of materials.

D.Use a laboratory galvanometer to make precise measurements.