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3 years ago

Approximately how long is a light-year?

Physics

2 answers:

Roman55 [17]3 years ago

8 0

C. is the correct response.

Burka [1]3 years ago

3 0

Answer:

1 light year= 9.5 trillion km

Explanation:

Light year is distance traveled by light in one year.

Distance = speed x time = speed of light x 1 year

1 light year= 3 x 10⁸ x 365 x 24 x 60 x 60 = 9.5 x 10¹⁵m=9.5x 10¹²km

1 light year= 9.5 trillion km

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the kinetic energy of an object of mass in moving with a velocity of 5 MS -1 is 25j what will be its kinetic energy when its vel

MAXImum [283]

Answer:

When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J

Explanation:

Kinetic Energy

Is the type of energy an object has due to its speed. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

The object has a kinetic energy of K=25 J when moving at v=5 m/s, thus the mass can be calculated by solving for m:

$\displaystyle m=\frac{2K}{v^2}$

$\displaystyle m=\frac{2*25}{5^2}=2$

m = 2 Kg

If the speed is doubled, v=10 m/s, the new kinetic energy is:

$\displaystyle K=\frac{1}{2}2\cdot 10^2$

K = 100 J

If the speed is tripled, v=15 m/s, the new kinetic energy is:

$\displaystyle K=\frac{1}{2}2\cdot 15^2$

K = 225 J

When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J

3 0

2 years ago

What is the kinetic energy of a 74.0 kg skydiver falling at a terminal velocity of 52.0m/s?

skad [1K]

Answer:

100,048

Explanation:

K.E = 1/2 m (v)^2

K.E = 1^/2 * 74 * (52)^2

K.E = 100,048J =100.048kJ

8 0

3 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$