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WITCHER [35]
4 years ago
5

A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r

otational kinetic energy
Physics
1 answer:
kykrilka [37]4 years ago
3 0

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = K.E=\frac{1}{2}I\omega^2

where,

I is the moment of inertia = mr²

on substituting the values, we get

K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2

or

K.E = 0.0075 J

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An ice cube tray full of water is put into a freezer. The water eventually reaches 0ºC and undergoes a phase change from a liqui
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"Darwin's finches" reside on the Galapagos Islands. Each species has a different type of beak, however, they can all be traced b
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Read 2 more answers
At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r
Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s    

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.                

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

W = \Delta Vq    (1)                                        

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C    

V: is the potential    

Also, the work is equal to:

W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}     (2)      

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

v_{a}: is the velocity in the point a

v_{b}: is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

\Delta Vq = \frac{1}{2}mv_{a}^{2}

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}

v_{a} = 2.9 \cdot 10^{4} m/s

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}

Since v_{a} = 0 (starts from rest) and m_{e} = 9.1x10⁻³¹ kg (electron's mass), we have:

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}    

v_{b} = 1.2 \cdot 10^{6} m/s

I hope it helps you!

6 0
4 years ago
A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele
HACTEHA [7]

Answer:

25 m

9.9 m/s

22 m/s

Explanation:

m = Mass of ball

v = Velocity

g = Acceleration due to gravity = 9.81 m/s²

Applying conservation of energy

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{g}\\\Rightarrow h=\dfrac{10^2}{2\times 9.81}\\\Rightarrow h=5.09683\ m

The height above the ground is 5.09683+20 = 25.09683 m = 25 m

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 5}\\\Rightarrow v=9.9\ m/s

The ball's speed as it passes the window on its way down is 9.9 m/s

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 25}\\\Rightarrow v=22.14723\ m/s

The speed of impact on the ground is 22 m/s

7 0
3 years ago
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