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Gennadij [26K]
4 years ago
12

A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releas

ing it at the bottom of the circle (underhand pitch). If the speed of the ball at the top of the circle was been 15 m/s, what will be the speed just after it's released?
Physics
1 answer:
Sever21 [200]4 years ago
3 0
Thank you for posting your question here at brainly. Below is the solution:

Ke up top = 1/2*.25 *225 
<span>gain of Pot energy = .25*9.81*1.2 </span>
<span>work input = (1/4)(2 pi *.6)*30 </span>

<span>so </span>
<span>sum of those 3 energies = </span>
<span>(1/2)(.25)v^2</span>
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Calculate the centripetal force exerted on the earth by the sun. Assume that the period of revolution for the earth is 365.25 da
Doss [256]

Answer:

F = 3.55 x 10²² N

Explanation:

average distance of sun from earth = 1.5 × 10⁸ km

mass of earth = 6 x 10²⁴ kg

time = 365.25 days

distance = 2 π r

distance = 2 π x 1.5 x 10¹¹ m

              = 3 π x 10¹¹ m

v = \dfrac{d}{t}

v = \dfrac{3 \pi \times 10^{11}}{365.25 \times 24 \times 60 \times 60}

   v = 2.98 x 10⁴ m/s

we know centripetal force

F = \dfrac{mv^2}{r}

F = \dfrac{6 \times 10^{24}\times (2.98\times 10^4)^2}{1.5\times 10^{11}}

F = 3.55 x 10²² N

8 0
3 years ago
The more _________ an electron has, the further away it can be from the nucleus. mass
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ectron's further from the nucleus are held more weakly by the nucleus, and thus can be removed by spending less energy. Hence we say they have higher energy. The Coulomb interaction energy between a nucleus of atomic number and an ele

4 0
4 years ago
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A crew team rows a boat at a rate of 20 km/h in still water. Beginning at time = 0 minutes, the team rows for 30 minutes up a ri
umka21 [38]
Since they are rowing against the current, their velocity realtive to the groud is 20-1.5 = 18.5km/h
So the distance travelled is 18.5km/h*0.5h = 9.25km
Going back with the current their ground realative velocity is 20+1.5 = 21.5km/h so the time it takes them to return 9.25 km is 9.25km/(21.5km/h) = 0.43h ~= 26 min
7 0
3 years ago
A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40
Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times  

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is \frac{1}{2}mv^2

So total energy at the top of the hill =mgh+0=mgh

At the bottom of the hill kinetic energy is equal to \frac{1}{2}mv^2 and potential energy will be 0

So total energy at the bottom of the hill is equal to 0+\frac{1}{2}mv^2

Form energy conservation mgh=\frac{1}{2}mv^2

v=\sqrt{2gh}, for v = 20 m/sec

20=\sqrt{2\times 9.8\times h}

Squaring both side

19.6h=400

h = 20.408 m

Now if velocity is 0 m/sec

40=\sqrt{2gh}

19.6h=1600

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

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Can anything block a magnet's force feild?
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Something that is not magnetic
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