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3 years ago

Which of kepler’s laws explains why the sun has a slightly larger angular diameter in january than in july?

1 answer:

7 0

Kepler’s three law is the answer. Kepler’s 3 is the amount of time it takes to orbit the sun is related to size and distance. Kepler’s 3 is one of the planetary motion and can be stated as all planets move in elliptical orbits, having the sun sits at one of the foci.

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A basketball has a coefficient of restitution of 0.821 in collisions with the wood floor of a basketball court. The ball is drop

Tanya [424]

Answer: The height of its fourth bounce = 0.43m

Explanation:

The coefficient of restitution denoted by (e), is the ratio that shows the final velocity to initial relative velocity between two objects after collision

IT is given by the formula in terms of height as

Coefficient of Restitution, e = √(2gh))/√(2gH) = √(h/H)

Where

Coefficient of Restitution, e= 0.821

H = 2.07 m

At fourth bounce , we have that

Coefficient of Restitution, e⁴ =√(h₄/H)

Putting the given values and solving , we have,

e⁴ =√(h₄/H)

= 0.821⁴ = √(h₄/2.07)

(0.821⁴ )² =h₄/2.07

0.2064 x 2.07 = 0.427 = 0.43

At fourth bounce, h₄ height = 0.43m

7 0

3 years ago

for every 120 joules of energy input a car waste 85 joules, find the useful energy output of the car?

harina [27]

Answer:

Useful Output = 35 J

Explanation:

The useful energy output of the car must be equal to the difference between the total input energy supplied to the car and the energy wasted by the car:

Useful Output = Total Input - Waste

where,

Total Input = 120 J

Waste = 85 J

Therefore,

Useful Output = 120 J - 85 J

<u>Useful Output = 35 J</u>

7 0

3 years ago

Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false?

marin [14]

Statements A, B, and C are true.

Statement <em> D is false</em>.

7 0

4 years ago

A wheel has a constant angular acceleration of 1.9 rad/s2. During a certain 7.0 s interval, it turns through an angle of 63 rad.

Anarel [89]

Answer:

4.5s

Explanation:

That must be the right answer.

3 0

3 years ago

A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/

Artyom0805 [142]

Answer:

$I=182.97\ kg-m^2$

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, $\tau=860\ N$

Angular acceleration of the shell, $\alpha =4.7\ m/s^2$

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

$\tau=I\alpha$

I is the rotational inertia of the shell

$I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2$

So, the rotational inertia of the shell is $182.97\ kg-m^2$ .

7 0

3 years ago