Answer:
Check the explanation
Explanation:
In calculate kinetic energy, write out a formula whereby the kinetic energy will be equal to 0.5 times mass times velocity squared. include in the value for the object mass, then the velocity with which it is moving.
Kindly check the attached image below to get the step by step explanation to the question above.
Answer:
7.98 m
Explanation:
In the given question,
distance above surface= 2 m
Distance penny from person = 8 m
Since the swimming pool is filled with water and atmosphere has air therefore the refractive index phenomenon will occur.
The refractive index of water: air is 4/3 (1.33).
Using the formula, 4/3 = real depth, apparent depth
real depth= 4/3 x apparent depth
Now, calculating apparent depth = 8 - 2
= 6 m
therefore, real depth = 4/3 x apparent depth
= 1.33 x 6
= 7.98
thus, 7.98 m is the real depth of water.
Answer:
Explanation:
For answer this we will use the law of the conservation of the angular momentum.
so:
where 
is the moment of inertia of the merry-go-round, 
is the initial angular velocity of the merry-go-round, 
is the moment of inertia of the merry-go-round and the child together and 
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where 
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:
Finally we replace all the data:
Solving for :
Answer:
Explanation:
Mass of ice m = 500g = .5 kg
Heat required to raise the temperature of ice by 10 degree
= mass of ice x specific heat of ice x change in temperature
= .5 x 2093 x 10 J
10465 J
Heat required to melt the ice
= mass of ice x latent heat
0.5 x 334 x 10³ J
167000 J
Heat required to raise its temperature to 18 degree
= mass x specific heat of water x rise in temperature
= .5 x 4182 x 18
=37638 J
Total heat
=10465 +167000+ 37638
=215103 J
Between the stars' absolute magnitudes<span> or </span>luminosities<span> versus their </span>stellar classifications<span> or </span>effective temperatures<span>. </span>
