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labwork [276]
3 years ago
11

Help!! HELP HELP HELP HELP

Physics
1 answer:
Anarel [89]3 years ago
8 0

Answer:

if a red <u>gaint </u> star is large enough, it may eventually blow up in an explosion called a <u>supernova</u><u> </u> leaving behind very dense <u>neutron</u> star.

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A toy car travels at 5.3 m/s. The car travels a distance of 17.8 m. How long did it
bonufazy [111]

Explanation:

Assuming constant speed:

Distance = speed × time

17.8 m = (5.3 m/s) t

t = 3.36 s

5 0
3 years ago
in as small room fan of rating 50 watt is used for 10hrs 2bulb of rating 10v are used for 8hrs daily.calculate monthly. electric
Pie

Answer:

The electric bill for June is Rs198000

Explanation:

Convert volt to watt, but in order to do so I need to know the amps and since it is not provided I converted if the amps was 1.

I multiple 50 with 10 then  with 30 so I know how much watt the fan takes at June.

Since there are 2 light bulb I multiple 10 with 2 than with 8 than with 30.

15000 watts for the fan,

4800 watts for light bulb,

add them and then times it by 10.

Rs198000

4 0
2 years ago
A Jack Rabbit hops 12.8 meters per second. How long would it take for him to hop 353 m? Round to the nearest tenth place. ​
adoni [48]

Time = Distance/speed

353/12.8 =27.57 seconds

4 0
2 years ago
AAAAAAAAAH
Dimas [21]

Answer:

l and lll

Explanation:

4 0
2 years ago
Read 2 more answers
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
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