To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:
Where,
m = Mass
v = Velocity
Replacing we have that the Total Kinetic Energy is
On the other hand the required Energy to heat up t melting point is
Where,
m = Mass
Specific Heat
Change at temperature
Latent heat of fussion
Heat required to heat up to melting point,
The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
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
Use stronger magnets
increase current
push magnets closer to coil
adding more sets of coils
Answer:
Increasing the speed of an object decreases its motion energy. Increasing the speed of an object increases its motion energy. Increasing the speed of an object does not affect its motion energy. Whether or not its motion energy is affected depends on how much its speed was increased.
Explanation:
