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Free_Kalibri [48]
3 years ago
7

A small ball is attached to one end of a massless, rigid rod. The ball and the rod revolve in a horizontal circle with the other

end of the rod at the center. The path of the ball has a constant linear speed. The force exerted by the rod is 0.5 N. The mass of the ball is 3kg. What is the centripetal acceleration of the ball, in m/s2?
Physics
1 answer:
nikklg [1K]3 years ago
6 0

Explanation:

<em>i </em><em>don't </em><em>have </em><em>keyboard </em><em>of </em><em>mathametics </em><em>so </em><em>i </em><em>solve </em><em>this</em>

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A satellite orbits at a distance from the Earth's center of about 2.60 Earth radii and takes 5.89 hours to go around once. What
Elenna [48]

Answer:

424088766.068 m

Explanation:

Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R

R = Radius of earth = 6371000 m (mean radius)

In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit

Circumference of a circle = 2×π×r

⇒Distance travelled in 5.89 hours = 2×π×2.6 R

⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000

⇒Distance travelled in 5.89 hours = 104078451.3393m

Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m

∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m

5 0
3 years ago
How many electrons can a nucleas hold
Hoochie [10]
A nuclear can hold 2 electrons
5 0
3 years ago
A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft
Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

m_1 v_1 = m_2 v_2

m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s

Using the equation to find v_1,

v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475

Using the conservation of energy equation, we have,

KE= \frac{1}{2}m*v^2

KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J

KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J

Total KE= 92077.75+13425530=92708.9J

Now this energy over the cannonball

KE=\frac{1}{2}m*v_2^2

92708.9=\frac{1}{2}15.5v_2^2

V_2 = 109.37m/s

The gain in velocity is 0.37m/s

4 0
3 years ago
A 4.0-m-diameter playground merry-go-round, with a moment of inertia of
HACTEHA [7]

Answer:

7.1 ms⁻¹

Explanation:

d = diameter of merry-go-round = 4 m

r = radius of merry-go-round = \frac{d}{2} =  \frac{4}{2} = 2 m

I = moment of inertia = 500 kgm²

w_{i} = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

w_{f} = angular velocity of merry-go-round after ryan jumps = 0 rad/s

v = velocity of ryan before jumping onto the merry-go-round

m = mass of ryan = 70 kg

Using conservation of angular momentum

Iw_{i} - m v r = (I + mr^{2})w_{f}

(500)(2.0) - (70) v (2) = (I + mr^{2})(0)

1000 = 140 v

v = 7.1 ms⁻¹

5 0
3 years ago
Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s
OleMash [197]

Answer:

My scenario would be A Car vs. a guard rail on a road.  You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

This is caused by so much force reacting from one object to another but also depends on molecular density.

5 0
3 years ago
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