A car is driving across a street. It's change in displacement is 50 m. If it started with a velocity of 5 m/s and ended with a v
1 answer:
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Answer:
a. Energy lost, Q = 16,800 Joules.
b. Power = 28 J/s
c. Time, t = 2357.14 seconds
d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
Explanation:
<u>Given the following data;</u>
- Mass = 0.20 kg
- Initial temperature, T1 = 20°C
- Final temperature = 0°C
- Time = 10 minutes
a. To find the energy lost by the water as it cools to 0 degree celsius;
Mathematically, heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- M represents the mass of an object.
- C represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 20 - 0
dt = 20°C
We know that the specific heat capacity of water is equal to 4200 J/kg°C
Substituting the values into the formula, we have;
<em>Energy lost, Q = 16,800 Joules.</em>
b. To find the average rate at which the water is losing energy in J/s by using the following formula;
First of all, we would have to convert the value of time in minutes to seconds.
<u>Conversion:</u>
1 minute = 60 seconds
10 minutes = X seconds
Cross-multiplying, we have;
X = 60 * 10
X = 600 seconds
Substituting the values into the formula, we have;
<em>Power = 28 J/s</em>
c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;
We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.
Mathematically, the latent heat of fusion is calculated by using the formula;
Energy, Q = ml = pt
Substituting the values into the formula, we have;
0.20 * 3.3 * 10⁵ = 28 * t
0.20 * 330000 = 28t
66000 = 28t
<em>Time, t = 2357.14 seconds.</em>
d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
Answer:
The mass of the ball is 0.23 kg
Explanation:
Given that
radius ,r= 3.74 cm
Density of the milk ,ρ = 1.04 g/cm³ = 1.04 x 10⁻³ kg/cm³
Normal force ,N= 9.03 x 10⁻² N
The volume of the ball V
V= 219.13 cm³
The bouncy force on the ball = Fb
Fb = ρ V g
Fb + N = m g
m=Mass of the ball = Density x volume
m = γ V , γ =Density of the Ball
ρ V g + N = γ V g ( take g= 10 m/s²)
γ = 0.00108 kg/cm³
m = γ V
m = 0.00108 x 219.13
m= 0.23 kg
The mass of the ball is 0.23 kg