1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.
<u>Explanation</u>:
We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the 
and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:
Better understood from numerical example as given:
If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?
This can be solved as follows:
It shows that man A will have more K.E.
Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.
Answer:
The centripetal acceleration for the first radius; 2.0 m = 50 m/s²
The centripetal acceleration for the second radius; 4.0 m = 25 m/s²
The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²
The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²
The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²
Explanation:
Given;
mass of the object, m = 1 kg
velocity of the object, v = 10 m/s
different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m
The centripetal acceleration for the first radius; 2.0 m
The centripetal acceleration for the second radius; 4.0 m
The centripetal acceleration for the third radius; 6.0 m
The centripetal acceleration for the fourth radius; 8.0 m
The centripetal acceleration for the fifth radius; 10.0 m
Answer:
serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
1 / Ceq = 1,147
Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
Dv = Q / Ceq
Q = DV Ceq
Q = 14 0.678 10⁻⁶
Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
ΔV = Q / C5
ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
ΔV = 1,725 V
d) The energy stores is
U = ½ C V²
U = ½ 0.678 10-6 14²
U = 66.4 10⁻⁶ J
e) Parallel system
Ceq = C1 + C2 + C3 + C4 + C5
Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
Q5 = C5 V
Q5 = 5.5 10⁻⁶ 14
Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
U = ½ C V²
U = ½ 22.7 10-6 14²
U = 2.2 10⁻³ J
Answer:
T₂ = 20.06 ° C
Explanation:
Given
P = 90 kg, T₁ = 20 ° C, h = 30 m, c = 1.82 kJ / Kg * ° C
Using the formula to determine the final temperature of the water
T₂ = T₁ * P * h / Eₐ * c
The work done of the person to the water
Eₐ = 1000 kg / m³ * 5 m³ * 9.8 m / s²
Eₐ = 49000 N
T₂ = 20 ° C +[ (90 kg * 30m) / (49000 N * 1.82) ]
T₂ = 20.06 ° C
Kinetic energy than parked
