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Crank
3 years ago
12

The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i

s a year on mercury in units of earth years?
Physics
1 answer:
Stels [109]3 years ago
5 0

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

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A certain light truck can go around a flat curve having a radius of 150 m with a maximum spee dof 32.0 m/s. With what maximum sp
Dvinal [7]

Answer

Maximum speed at 75 m radius will be 22.625 m /sec

Explanation:              

We have given radius of the curve r = 150 m

Maximum speed v_{max}=32m/sec

Coefficient of friction \mu =\frac{v_{max}^2}{rg}=\frac{32^2}{150\times 9.8}=0.6965

Now new radius r = 75 m

So maximum speed at new radius v_{max}=\sqrt{\mu rg}=\sqrt{0.6965\times 75\times 9.8}=22.625m/sec

7 0
3 years ago
At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw
allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

f_i = f_0 \frac{c}{c+v}

Where

f_h= Approaching velocities

f_i= Receding velocities

c = Speed of sound

v = Emitter speed

And

f_h = f_0 \frac{c}{c+v}

Therefore using the values given we can find the velocity through,

\frac{f_h}{f_0}=\frac{c-v}{c+v}

v = c(\frac{f_h-f_i}{f_h+f_i})

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

v = 353(\frac{2.4-1}{2.4+1})

v = 145.35m/s

Therefore the cars goes to 145.3m/s

7 0
3 years ago
An object is observed for a time interval of 20 seconds. From time 6.7 s the object experiences a Force of 106 N that lasts unti
shusha [124]

Answer:

1654 kg m/s

Explanation:

The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:

I=F\Delta t

where:

I is the impulse

F is the force exerted on the object

\Delta t is the time during which the force is applied

For the object in this problem, we have

F=106 N (force applied)

\Delta t= 15.6 s (time interval)

Therefore, the impulse experienced by the object is:

I=(106)(15.6)=1654 kg m/s

3 0
3 years ago
ASAP WILL GIVE BRAINLIEST!!!
Oksana_A [137]

Answer:

<em>The change in momentum of the car is 3575 Kg.m/s</em>

Explanation:

<u>Impulse and Momentum</u>

The impulse (J) experienced by the object equals the change in momentum of the object (Δp).

The formula that represents the above statement is:

J = Δp

The impulse is calculated as

J = F.t

Where F is the applied force and t is the time.

The car hits a wall with a force of F=6500 N and stops in 0.55 s. Thus, the impulse is:

J = 6500 * 0.55

J = 3575 Kg.m/s

The change in momentum of the car is:

\Delta p= J = 3575\ Kg.m/s

The change in momentum of the car is 3575 Kg.m/s

5 0
3 years ago
Michael's car ran over a nail, which made a hole in one tire. at what point will air stop pouring out of the hole in the tire.​
aniked [119]
When the tire has released all of its pressure OR when it runs out of air
6 0
3 years ago
Read 2 more answers
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