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kirza4 [7]
3 years ago
10

Which of these travels the fastest?

Physics
1 answer:
bogdanovich [222]3 years ago
8 0
C) Light in vacuum

Light in a vacuum, on the order of magnitudes, travels much faster than the other three. In fact, the speed of light in a vacuum is the theoretical limit of how fast anything can travel. 
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To practice problem-solving strategy 26.1 resistors in series and parallel. two bulbs are connected in parallel across a source
hichkok12 [17]
Emf e = 11
r 1 = 3.0
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r 3 = ?

The two in parallel are equivalent to 3 • 3/6 = 1.5 Ω 
To have 2.4 volts across them, the current is I = 2.4/1.5 = 1.6 amps. and the unknown R = (11–2.4) / 1.6 = 5.375 Ω or 5.4 Ω 
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3 years ago
Please check my answers!
Alecsey [184]
A.) The higher the altitude, the colder the climate will be

B.) Areas near the equator have warmer climates than areas for form the equator.

D.) Winds that blow inland from oceans or large lakes contain a lot of water vapor that will cause precipitation.

C.) Monsoons.
8 0
3 years ago
Read 2 more answers
Three-point charges are arranged on a line. Charge q3 = +5.00nC and is at the origin. Charge q2 = -3.00 nC and is at x = 5.00 cm
Anon25 [30]

Answer:

Explanation:

Given

Charge q_3=+5\ nC is placed at origin

Charge q_2=-3\ nC is placed at x=5\ cm

Charge q_1 is Placed at x=2.5\ cm

charge q_1 must be positive so as to balance the force on charge q_3

Force on q_3 due to q_1 is

F_{31}=\frac{kq_1q_3}{r^2}

here r=2.5\ cm

F_{31}=\frac{kq_1q_3}{(2.5)^2}

Force on q_3 due to q_2 is

F_{32}=\frac{kq_3q_2}{r^2}

here r=5\ cm

F_{32}=\frac{kq_3q_2}{(5)^2}

F_{31}=F_{32}

\frac{kq_1q_3}{(2.5)^2}=\frac{kq_3q_2}{(5)^2}

q_1=q_2(\frac{2.5}{5})^2

q_1=3\times (\frac{2.5}{5})^2

q_1=3\times \frac{1}{4}

q_1=0.75\ nC

                                                                   

5 0
4 years ago
Julia and her musician friends were competing in the school talent show. Julia wanted her band to win the "most talented" award,
jeyben [28]

No, because she is hoping people vote for her.

4 0
3 years ago
Read 2 more answers
300 grams of ethanol is heated with 14640 Joules of energy to reach a final temperature of 30 °C. What was the initial temperatu
levacccp [35]

Answer:

10 °C

Explanation:

Applying

q = cm(t₂-t₁)............... Equation 2

Where q = heat energy, c = specific heat of ethanol, m = mass of ethanol, t₁ = initial temperature, t₂ = Final temperature.

Given: c = 2.44 J/g.°C,  m = 300 g, q = 14640 J, t₂ = 30°C

Substitute into equation 2 and solve for t₁

14640 = 2.44×300(30-t₁)

14640 = 732(30-t₁)

732(30-t₁) = 14640

(30-t₁)  = 14640/732

(30-t₁)  = 20

t₁ = 30-20

t₁ = 10 °C

6 0
3 years ago
Read 2 more answers
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