When you heat a certain substance with a difference of temperature

the heat (energy) you must give to it is

where

is the specific heat of that substance (given in J/(g*Celsius))
In this case

Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin) because on the temperature scale a
difference of 1 degree Celsius = 1 degree Kelvin
Weight is different (but mass is the same)
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
<u>T₂ = 95.56°C</u>
Answer:
Induced emf, 
Explanation:
Given that,
Length of the helicopter, l = 4 m
Angular speed of the helicopter, 
The vertical component of the Earth’s magnetic field is, 
We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :



So, the induced emf between the tip of a blade and the hub is
. Hence, this is the required solution.
Answer:The net force on the block is zero.
Explanation:
Given
Block is being pulled upward along an inclined surface at a constant speed
As speed is constant and moved in a straight line along the plane therefore its velocity is also constant .
and change in velocity is equal to acceleration therefore acceleration is zero here i.e. net force is zero acting on the body.