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Oduvanchick [21]
3 years ago
14

how do u calculate the kinetic energy of a ball of mass 0.25kg being kicked vertically upwards with a speed of 5m/s​

Physics
1 answer:
vfiekz [6]3 years ago
6 0

Answer:

3.125J

Explanation:

K.E.= 1/2(mass)(velocity)^2

K.E.=1/2(0.25)(5)^2=3.125

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How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams
Harrizon [31]

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water  = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation  = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

         H  = mL

m is the mass

L is the latent heat of vaporization

Insert the parameters and solve;

         H  = 20g x 2260J/g

          H = 45200J

4 0
2 years ago
What is it called when a surface takes light in without reflecting it
andrew-mc [135]
Reflection from such a rough surface is called diffuse reflection and appears matte
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3 years ago
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The chicken's eggs are fertilized interally or externally
zlopas [31]
The chicken's eggs are fertilized interally
6 0
3 years ago
Why would knowing the characteristics of circuits be important in designing electrical circuits?
zubka84 [21]
Well if you didn't you could make mistakes, which would lead ,in the best case, at a fail of the circuit , or if it goes out of control it could be dangerous

for example you have to know that the wires become hot and they loose their abbilitys as connecters(the hotter it will, the more energy you lose becouse the R will be bigger)
8 0
3 years ago
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A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo
CaHeK987 [17]

Answer:\omega =93.51 rad/s

Explanation:

Given

mass of Flywheel m_1=1440 kg

mass of bus m_b=10200 kg

radius of Flywheel r=0.63 m

final speed of bus v=21 m/s

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let \omegabe the angular velocity of Flywheel

0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}

I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2

0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}

\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}

\omega =21\times 4.45=93.51 rad/s

8 0
3 years ago
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