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KATRIN_1 [288]
4 years ago
5

PHYSICAL SCIENCE HELP PLEASE !!!!! Picture below is for question 4

Physics
1 answer:
Eva8 [605]4 years ago
3 0

Question 1) excited and moving around a lot.

Question 2) heat

Question 3) specific heat

Question 4) As temperature increases, energy transferred increases.

Hope these answers help!

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1. If you add air to a flat tire through a single small entry hole, why does the air spread out to fill the tire?
ikadub [295]

Answer:

The tire fills up just like anything else that holds air when u pump a ball or tire up it fills all the way up cause it is a small confined space and after filling it with air the atoms of the air fill the tire up

Explanation:

7 0
3 years ago
1.) Using Ohm’s law, explain how voltage changes in relation to current, assuming that resistance remains constant.
Harrizon [31]
1.) Ohm's law is understood as I = V/R. Given that resistance is constant, then voltage changes directly proportional to current.

2.) The more current that passes through a lightbulb, the brighter it glows. The higher the current, the higher the power, where power determines the brightness of a bulb.

3.) A bulb has a specific limit to how much power (Watts) it can handle. Going over the limit would cause the bulb to burn out.

4.) When a bulb burns out, no current will be able to pass through the filament.
4 0
3 years ago
Read 2 more answers
Describe how the particles of a substance behave at its boiling point.
fgiga [73]
Particles begin to move out of their ordered arrangement due to energy. They are unable to escape because of external pressure pushing down on the liquid.
4 0
3 years ago
Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h
vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

R_A=2R_B

Also Potential of both sphere is same

V_A=V_B

V=\frac{kQ}{R}

thus

k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}

\frac{Q_A}{Q_B}=\frac{R_A}{R_B}

\frac{Q_A}{Q_B}=\frac{2}{1}=2

\frac{Q_B}{Q_A}=\frac{1}{2}

(b)Ratio of \frac{E_B}{E_A}

Electric Field is given by E=\frac{kQ}{R^2}

thus E_A=\frac{kQ_A}{R_A^2}----1

E_B=\frac{kQ_B}{R_B^2}----2

Divide 2 by 1

\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}

\frac{E_B}{E_A}=\frac{1}{2}\times 4=2

8 0
3 years ago
an 11 ohm resistor and a 6 ohm resistor are connected in series with a battery. the potential difference across the 6 ohm resist
Maru [420]

Two resistors in series are often called a 'voltage divider', because the
total voltage divides in proportion to the resistances.

The total resistance in the string across the battery is (11 + 6) = 17 ohms.

-- The full battery voltage appears across 17 ohms.
-- The voltage across the 11-ohms is (11/17) of the battery, and
-- the voltage across the 6-ohms is (6/17) of the battery. 

                                  (6/17) x (B) = 9 volts

Multiply each side by (17/6) :      B = (9 volts) x (17/6)  =<em>  25.5 volts </em>.

By the way, in case you care or are asked . . .

-- The current in the whole series loop is  B/R = 25.5 / 17 = 1.5 Amperes
-- The power drawn from the battery is   B²/R = (25.5)²/17 =  38.25 watts
-- The power dissipated by the 6-ohm resistor is  V²/R = 9²/6 = 13.5 watts
-- The power dissipated by the 11-ohm resistor is I²R = (1.5)² (11) = 24.75W
-- (Check:  13.5W + 24.75W = 38.25W     yay! )
-- If they're just composition units hanging out in the air, then both resistors
are getting quite warm.


4 0
3 years ago
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