Add the KE increase and the work done against friction.
The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J
The friction work done is 6*3.8 = 22.8 J
hope this is correct
Answer:
(a) 5.43 x 10⁵ J
(b) 3.07 x 10⁵ J
(c) 45 °C
Explanation:
(a)
= Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg
m = mass of ice = 1.63 kg
= Energy required to melt the ice
Energy required to melt the ice is given as
= m
= (1.63) (3.33 x 10⁵)
= 5.43 x 10⁵ J
(b)
E = Total energy transferred = 8.50 x 10⁵ J
Q = Amount of energy remaining to raise the temperature
Using conservation of energy
E =
+ Q
8.50 x 10⁵ = 5.43 x 10⁵ + Q
Q = 3.07 x 10⁵ J
(c)
T₀ = initial temperature = 0°C
T = Final temperature
m = mass of water = 1.63 kg
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J
Using the equation
Q = m c (T - T₀)
3.07 x 10⁵ = (1.63) (4186) (T - 0)
T = 45 °C
The specific gravity of the object’s material is 5.09.
<h3>To calculate the specific gravity of the object:</h3>
Weight difference = 9 - 7.2 = 1.8 N = Buoyant force of water
Buoyant Force in water(Fb) = density of water x g x volume of the body(Vb)
1.8 = 1000 x 9.81 x Vb
Vb = 1.8/9810 cubic meter
Now, in the air;
Weight of body = mg = 9 N
Mass of body,m = 9/9.81 Kg
So,
Density of body = m/ Vb
= 9/9.81 ÷ 1.8/9810
= 5094.44 kg per cubic meter
The specific gravity of body = density of body ÷ density of water
= 5094.44 ÷ 1000
= 5.09
Therefore, Specific gravity of body = 5.09
Learn more about Specific gravity here:
brainly.com/question/13258933
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<span>We use m/s for the velocity, and m2/(V·s) for the </span><span>mobility.</span>
They may be used whenever you are not experienced with swimming and you are near the water such as swimming, canoeing, etc