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Taya2010 [7]
2 years ago
11

The asteroid ceres lies at an average distance of 414 million kilometers from the sun. The period of revolution of ceres around

the sun is approximately

Physics
2 answers:
nikitadnepr [17]2 years ago
8 0

Answer:

Answer is 1.156x10^23 sec

Explanation:

According to kepler's law, the square of the period of revolution is proportional to the cube of the distance within planets.

At a distance r of 414 million kilometers I.e 414x10^6 km, period calculation is shown in the image below.

7nadin3 [17]2 years ago
6 0

Answer:

T2 = 1680,4 days

Explanation:

Kepplers law:

\frac{T^{2} }{a^{3} }  = constant

For Earth:

T1 = 365 days ; a1 = 149 597 870 700 m

For Ceres:

T2 = ? days ; a1 = 414*10^{9}  m

Then:

\frac{T1^{2} }{a1^{3} }  = \frac{T2^{2} }{a2^{3} }  ----> T2 =  T1*\sqrt{\frac{a2^{3}}{a1^{3}}}

Replacing values:

T2 = 1680,4 days

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A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
never [62]

Answer:

1) Vf = 3.36 m/s

2) v = 6.86 m/s

3) s = 92.3 m

4) a = - 0.23 m/s²

Explanation:

1)

We use first equation of motion in this case:

Vf = Vi + at

where,

Vf = Final velocity = ?

Vi = Initial velocity = 0 m/s

a = acceleration = 1.4 m/s²

t = time = 2.4 s

Therefore,

Vf = 0 m/s + (1.4 m/s²)(2.4 s)

<u>Vf = 3.36 m/s</u>

2)

We again use first equation of motion but with t= 4.9 s now:

Vf = 0 m/s + (1.4 m/s²)(4.9 s)

Vf = 6.86 m/s

Now, this velocity remained constant for net 11 seconds. Hence, the velocity of hare after 8.9 s is:

<u>v = 6.86 m/s</u>

3)

First we use second equation of motion to find distance covered in accelerated motion:

s₁ = Vi t + (0.5)at²

s₁ = (0 m/s)(4.9 s) + (0.5)(1.4 m/s²)(4.9 s)²

s₁ = 16.8 m

Now, we calculate the distance covered in uniform motion:

s₂ = vt

s₂ = (6.86 m/s)(11 s)

s₂ = 75.5 m

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s = s₁ + s₂ = 16.8 m + 75.5 m

<u>s = 92.3 m</u>

3)

Using third equation of motion for the decelerated motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance covered = 102 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = 6.86 m/s

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(2)(a)(102 m) = (0 m/s)² - (6.86 m/s)²

a = - (47.06 m²/s²)/(204 m)

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3 years ago
What simple machine makes up most of the joints in your body
Vladimir [108]
Levers
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3 0
2 years ago
You're in your room blasting music with door shut, your mom opens your door. Now music is heard through out your home. this is e
Sliva [168]

Answer:

A

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5 0
2 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

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\theta is the angular displacement of the object

t is the time elapsed

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In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

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t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

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2)

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v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

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h = 600 km

over the Earth's surface, which has a radius of

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