<u>Thermal energy</u><u> from the room-temperature water will continuously flow to the boiling water.</u>
- The second law states, in a straightforward manner, that heat cannot naturally go "uphill."
- When a pan of boiling water and a pan of ice are in touch, the hot water cools and the ice melts and warms up.
<h3>
THE FIRST LAW OF THERMODYNAMICS</h3>
- Adiabatic Process - is a procedure that is carried out without the system's heat content changing.
- Water is heated to a temperature of 1000C during the boiling process, making it an isothermal process. As steam, the excess heat leaves the system.
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Answer:
Explanation:
Given that,
Number of turn is 48
N=48
Radius is 4.8cm
r=0.048m
Magnetic Field
B=0.48T
Current in coil
i=23.3mA
i=0.233A
Maximum Torque?
Maximum torque occur at angle 90°
Torque is given as
τ = N•I•A•B•sinθ
Where N is number of turn =48
I is current in coil =0.233A
A is area of circular coil form
Area of a circle is given as
A=πr²
A=π×0.048²
A=0.007238m²
B is magnetic field =0.48T
Maximum torque occurs at 90°
τ = N•I•A•B•sinθ
τ=48×0.233×0.007238×0.48×Sin90
τ = 0.0389Nm
This torque is large enough to exert the coil
Answer:
All the answers are solved and explained below.
Explanation:
Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.
Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).
Data Given:
m = 1000 kg
F = 2000N
t = 10s
Q1 Solution:
Acceleration = a = ?
F = ma
a = F/m
a = 2000/ 1000
a = 2 
Q2: Solution:
Change in velocity = Δv = ?
acceleration = change in velocity / time
a = Δv/t
Δv = axt
Δv = 2 x 10
Δv = 20 m/s
Q3: Solution:
Impulse = I = ?
Impulse = Force x time
I = 2000 x 10
I = 20000 Ns
Q4: Solution:
Change in Momentum = Δp = ?
Δp = mΔv
Δp = 1000 x 20
Δp = 20000 Kgm/s
Q5: Solution:
Final velocity of the car at the end of 10 seconds = vf = ?
Δp = m x Δv
Δp = m x (vf-vi)
Δp = 1000 x (vf - 0 )
20000 = 1000 x vf
vf = 20000/1000
vf = 20 m/s
Q6: Solution:
Change in momentum the car experiences as it continues at this velocity?
Δp = ?
Δp = mΔv
Δp = m x (0)
Δp = 0
Q7: Solution:
Impulse = Change in momentum
Impulse = Δp
Implulse = 0
Q8: Solution:
Change in momentum = Δp = mΔv
Δp = m(vf-vi)
Δp = 1000 x (0-20)
Δp = -20000 kgm/s
Q9: Solution:
Impulse = Δp
Impulse = -20000 Ns
Q10: Solution:
Impulse = ?
Impulse = F x t
F = impulse/t
F = -20000/4s
F = -5000 N
Q11: Solution:
F = ma
a = ?
a = F/m
a = -5000/1000
a = -5