Answer:
V(c)' = 0.325 m/s
V(a)' = 2.19375 m/s
V(b)'' = 0.73125
Explanation:
See attachment for explanation.
When current is passed through a hot plate, it converts _______ energy into ________ energy
2 answers:
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Answer:
(1) 2.05 x 10^4 N/C
(2) Downward
(3) upward, 9.8 m/s^2
(4) upward, 9.8 m/s^2
Explanation:
q = - 2.1 micro coulomb, m = 0.0044 kg, g = 9.8 m/s^2
(1) The electric force is given by F = - q x E
The magnitude of electric force is balanced by the weight of the charged particle
q x E = m x g
E = mg / q
E = 2.05 x 10^4 N/C
(2) As the electric force is acting upward and the weight is downward so the elecric field is in downward direction.
(3) The charge is doubled,
then the electric field becomes half.
E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C
The direction is same that is in downward direction.
Acceleration = Force / mass
a = mg / m = 9.8 m/s^2 upward
(4) The charge is doubled,
then the electric field becomes half.
E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C
The direction is same that is in downward direction.
Acceleration = Force / mass
a = mg / m = 9.8 m/s^2 upward
The energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
U = m ΔT
U = Energy
= Specific heat
m = Mass
ΔT = Change in temperature
ρ = Density
V = Volume
ρ = 1000 g / m³ (Dry air )
= 1 J / g K
ΔT = 5 °C
V = 4 * 4 * 3
V = 48 m³
m = ρ V
m = 1000 * 48
m = 48000 g
U = 1 * 48000 * 7
U = 336000 J
U = 336 kJ
Therefore, the energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
To know more about energy required to raise temperature
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