Complete Question:
Check the circuit in the file attached to this solution
Answer:
Total current = 0.056 A(From left to right)
Explanation:
Let the current in loop 1 be I₁ and the current in loop 2 be I₂
Applying KVL to loop 1
30 - (I₁ - I₂)500 + I₂R + 15 = 0
45 - 500I₁ - 500I₂ + RI₂ = 0
I₁ = 30mA = 0.03 A
45 - 500(0.03) - 500I₂ + RI₂ = 0
30 -500I₂ + RI₂ = 0...............(1)
Applying kvl to loop 2
-RI₂ - 15 + 10 - 400I₁ = 0
-RI₂ = 5 + 400*0.03
RI₂ = -17 ................(2)
Put equation (2) into (1)
30 -500I₂ -17 = 0
-500I₂ = 13
I₂ = -13/500
I₂ = -0.026 A
The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026
Total current = 0.056 A
The current will flow from left to right
I don't understand what you mean by "depth" of the steps. The flat part of the step has a front-to-back dimension, and the 'riser' has a height. I don't care about the horizontal dimension of the step because it doesn't add anything to the climber's potential energy. And if the riser of each step is 20cm high, then 3,234 of them only take him (3,234 x 0.2) = 646.8 meters up off the ground. So something is definitely fishy about the steps.
Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.
In order to lift his bulk 828 meters from the street to the top of the Burj, the climber has to provide a force of 800 newtons, and maintain it through a distance of 828 meters. The work [s]he does is (force) x (distance) = <em>662,400 joules. </em>
Answer:
Explanation:
Mass of a hockey puck, m = 0.17 kg
Force exerted by the hockey puck, F' = 35 N
The force of friction, f = 2.7 N
We need to find the acceleration of the hockey puck.
Net force, F=F'-f
F=35-2.7
F=32.3 N
Now, using second law of motion,
F = ma
a is the acceleration of the hockey puck
So, the acceleration of the hockey puck is 
.
Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
i = 3.4 A
(c)
Current density is given as
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
E = 9.01 x 10⁻³ V/m
