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4 years ago

What do kepplers laws describe?

1 answer:

6 0

Kepler's laws describe how the planets move in an orbit around the sun, so the answer would be C. The shape of an orbit.

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A 7.26kg bowling ball (16 pounds) is at rest what is the net force on the bowling ball

cestrela7 [59]

Well since the bowling ball is at rest and is not moving at all, and assuming that the ball is on an even surface, the Fg is equal to the FN = normal force, and thus all forces are balanced, the Fnet = 0.

5 0

3 years ago

How do you find out the missing masses in a balloon

Mamont248 [21]

Answer:

How do you find out the missing masses in a balloon?

Well, actually you can't find the missing masses in a balloon. Why?

Because the mass of the mass of balloon, it can't see the mass of it, it only see if it the balloon is deflated or inflated.

Explanation:

Hope it helps

#LetsStudy

7 0

3 years ago

Figure 23.9 shows a sliding mass on a spring. Assume there is no friction.

Tom [10]

<span>Without friction, there will be undamped simple harmonic motion. The force of the spring is proportional to the distance from the equilibrium point. The period of oscillation will be independent of the amplitude.

I hope my answer has come to your help. God bless and have a nice day ahead!</span>

4 0

3 years ago

Abox has a mass of 14.4 kg. The length is 4 m longthe width is 1 m and the height is 5 m. What is the density of the box?

valkas [14]

Answer:

0.72 kg per cubic m

Explanation:

Mass = 14.4 kg

Volume = lbh = 4*1*5 = 20 cubic m

$\because \: density \\ \\ = \frac{mass}{volume} \\ \\ = \frac{14.4}{20} \\ \\ = 0.72 \: kg {m}^{ - 3}$

3 0

3 years ago

A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.

N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

$\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2$

(c) Moment of inertia :

The net torque acting on it is, $\tau=I\alpha$ , I is the moment of inertia

Also, $\tau=Fr$

So,

$I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2$

Hence, this is the required solution.

3 0

3 years ago