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Arada [10]
3 years ago
6

Radio waves travel at the speed of light, which is 3.00 u 108 m/s. How many kilometers will radio messages travel in exactly one

year?
Physics
2 answers:
egoroff_w [7]3 years ago
8 0

Answer:                

Distance traveled will be 9.462\times 10^{12}km

Explanation:

We have given speed of the light v=3\times 10^8m/sec

We have to find the distance traveled by light in 1 year

We know that 1 year = 8760 hour

And 1 hour = 60×60 = 3600 sec

So 1 year =8760\times 3600=3.154\times 10^{7}sec

We know that distance = speed × time

So distance traveled by light in one year =3.154\times 10^7\times 3\times 10^8=9.462\times 10^{15}m

We have to fond the distance in km

As we know that 1 km = 1000 m

So 9.462\times 10^{15}m=\frac{9.462\times 10^{15}}{1000}=9.462\times 10^{12}km

Scrat [10]3 years ago
3 0

Answer:

9.46 x 10^15 m

Explanation:

speed, v = 3 x 10^8 m/s

time, t = 1 year = 1 x 365 x 24 x 3600 s = 31536000 s

distance = velocity x time

d = 3 x 10^8 x 31536000

d = 9.46 x 10^15 m

Thus, the distance is 9.46 x 10^15 m.

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A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is
Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
P = i^2R

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

P = (2.0^2)(50) = \boxed{\text{ B. }200 W}

7 0
2 years ago
A 1.00-kg object is attached by a thread of negligible mass, which passes over a pulley of negligible mass, to a 2.00-kg object.
Anna [14]

Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

given,

mass A = 1 kg

mass B = 2 kg

vertical distance between them = 1 m

F_d = mg

F_d = 2 \times 9.8

F_d = 19.6\ N

F_u = mg

F_u = 1 \times 9.8

F_u = 9.8\ N

F_{net} = 19.6 - 9.8

F_{net}=9.8\ N

F = (m_1+m_2)a

9.8 = (2+1)a

a = 3.27 m/s²

The speed of the system at that moment is:

v² = u² + 2×a×s

v² = 0² + 2× 3.27 × 1

v ² = 6.54

v = 2.56 m/s

3 0
3 years ago
A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan
ASHA 777 [7]

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of a \;\;m/s^2 from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, a=\frac{v-u}{t}=\frac{5-4.2}{t}

\Rightarrow at=0.8\cdots(i)

Now, from the equation of motion

s=ut+\frac 12 at^2

\Rightarrow s=ut+\frac 12 at(t)

\Rightarrow 115=4.2t+\frac 12 \times 0.8 t [ from equation (i)]

\Rightarrow 115=(4.2+0.4)t

\Rightarrow t= 115/4.6 = 25 seconds.

Hence, she takes 25 seconds to walk the distance.

5 0
3 years ago
Derive the explicit rule for the pattern <br> 3, 0, -3, -6, -9,
pickupchik [31]
It is the last number minus 3
8 0
3 years ago
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A wagon is accelerating down a hill. Which statement is true?
Ira Lisetskai [31]
Potential energy decreases and kinetic energy increases.

Potential energy is related to the height, since the wagon is going downhill, height decreases and potential energy decreases.

Kinetic energy is related to the speed, since the wagon is speeding up, kinetic energy increases.
3 0
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