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Rufina [12.5K]
3 years ago
15

Use the worked example above to help you solve this problem. A car traveling at a constant speed of 27.9 m/s passes a trooper hi

dden behind a billboard, as in the figure. One second after the speeding car passes the billboard, the trooper sets off in chase with a constant acceleration of 3.36 m/s2.
(a) How long does it take the trooper to overtake the speeding car?
Physics
1 answer:
Sidana [21]3 years ago
7 0
H h h huh h jvjvhvuvuvuvyv
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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
When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is m. When the mass of the bottle is 0.250 kg
Jet001 [13]

Answer:

when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.

when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.

when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.

when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.

Explanation:

5 0
3 years ago
If the distance between two asteroids is doubled, the gravitational force they exert on each other will.
True [87]
The force will be 4 times smaller.
5 0
3 years ago
If the sum of all the forces acting on a car is zero, then the speed of the car?
Taya2010 [7]

Answer: If all forces acting on a car are zero, than the cars speed is zero since there are no forces to push or pull the car :)

Explanation:

7 0
2 years ago
Calculate the kinetic energy in joules of a ball of mass 40g moving at a velocity of 4 metres per second​
AfilCa [17]
Given : A ball of mass 40 g moving at a velocity of 4 m/s.
To find : Calculate the kinetic energy in joules ?
Solution :
The kinetic energy formula is given by,

where, v is the velocity v=4 m/s
m is the mass m=40 g
Convert g into kg,



Substitute the values,



Therefore, the kinetic energy is 0.32 Joules.
6 0
2 years ago
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