(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.
(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.
<h3>
Speed of the block when pushed by the spring</h3>
The speed of the block when pushed by the spring is calculated as follows;
K.E = Ux
¹/₂mv² = ¹/₂kx²
mv² = kx²
v² = kx²/m
v² = (25 x 0.5²)/0.05
v² = 125
v = 11.18 m/s
<h3>Final velocity of the two balls after the collision</h3>
The velocity of the two balls after the collision is calculated as follows;
Pi = Pf
where;
- Pi is initial momentum
- Pf is final momentum
m1u1 + m2u2 = v(m1 + m2)
0.05(11.18) + 0.05(0) = v(0.05 + 0.05)
0.559 = 0.1v
v = 5.59 m/s
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Answer:
C
Explanation:
First the water heats up to the boiling point ( temp increases)
then, as it boils it remains at constant temp ( boiling point)
Answer: 37.5 kg in 3 s.f.
Explanation:
Answer:
0.0133A
Explanation:
Since we have two sections, for the Inductor region there would be a current
. In the case of resistance 2, it will cross a current
Defined this we proceed to obtain our equations,
For
,


For
,


The current in the entire battery is equivalent to,


Our values are,




Replacing in the current for t= 0.4m/s



<span>3933 watts
At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3
The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams.
Rounding to 4 significant figures gives me 4705 grams of water.
The heat of vaporization for water is 2257 J/g. So the total energy applied is
2257 J/g * 4705 g = 10619185 J
Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds.
Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So
10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3)
= 3933 watts</span>