Explanation:
It is given that,
An electron is released from rest in a weak electric field of, 
Vertical distance covered, 
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :
.............(1)
Electric force is 
and force of gravity is 
. As both forces are acting in downward direction. So, total force is:
Acceleration of the electron, 
Put the value of a in equation (1) as :
v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.
So they give us this
V=IR
V= 1.8
I=0.4
R=?
So we insert the thing that we know.
1.8=0.4*R
We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.
So we have this.
Lastly we solve.
R=4.5ohms
The formula to find R is V=IR
V/I=R
So the resistance will be the Voltage divided by the Current
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
I believe it’s C. Plane .
Resistance = Voltage/Current
Wattage = Voltage * Current
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That means the current drawn by the lamp is equal to 100 watts divided by 210 volts.
Resistance =
