Answer:
the speed and aceleration of the ping pong ball is greater than that of the bowling ball.
Explanation:
We can analyze this exercise from several points of view, if we use Newton's second law
Bowling ball
F = M a₁
pingpongg ball
F = m a₂
as the forces the same
M a₁ = m a₂
a₂ =
a₁
Since the mass of the bowling ball is much greater than the ping pong ball,
a₂ »a₁
so the acceleration of the ping pong ball is much greater than the acceleration of the bowling ball.
If we use the relationship of momentum and momentum, assuming that the time for the two cases is the same and that both start from rest
Bowling ball
I = F t = Δp
I = M (v₁ - v₀)
Ping pong ball
I = F t = Δp
I = m (v₂ -v₀)
the impulse itself
M v₁ = m v₂
v₂ =
v₁
so we conclude that the speed of the ping pong ball is much greater than the speed of the bowling ball.
In conclusion the speed and aceleration of the ping pong ball is greater than that of the bowling ball.
General waste - medical waste that does not pose any harmful threats
Hazardous waste - medical waste that can be harmful but does not pose a risk to infection
Infectious waste - medical waste materials that poses risk of infection to humans
Radioactive waste - medical waste that contains radioactive material
Answer: The angular speed w of the loop = 181 rad/s
Explanation:
Given that;
Area A = 0.193 m2
Magnetic field B = 0.374 T
E.m.f = 9.24v
Ø = 45 degree
According to Faraday's law when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.
Using the formula
E.m.f = NABwcosØ
Where w = angular velocity.
Let assume that N = 1 then,
9.24 = 0.193 × 0.374 × w × cos45
9.24 = 0.051w
w = 9.24/0.051
w = 181 m/s
Therefore, the angular speed w of the loop = 181 rad/s
Explanation:
The given data is as follows.
mass, m = 75 g


Specific heat of water = 4.18
First, we will calculate the heat required for water is as follows.
q = 
= 
= 8464.5 J/mol
= 8.46 kJ ......... (1)
Also, it is given that
= (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.
Now, we will calculate the heat of fusion as follows.
q = 
= 
= -46.32 kJ ......... (2)
Now, adding both equations (1) and (2) as follows.
8.46 kJ - 46.32 kJ
= -37.86 kJ
Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at
to ice at
is -37.86 kJ.