Answer:
Therefore,
Current through Nichrome wire is 0.3879 Ampere.
Explanation:
Given:
Length = l = 10 meter
V = 12 Volt
To Find:
Current, I =?
Solution:
Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as
Where,
R = Resistance
l = length
A = Area of cross section = πr²
Substituting the values we get
Now by Ohm's Law,
Substituting the values we get
Therefore,
Current through Nichrome wire is 0.3879 Ampere.
Answer:
1. The force of the shelf holding the book up.
Explanation:
The free body diagram of the book is as follows:
1 - The weight of the book towards downwards
2 - The normal force that the shelf exerts on the book towards upwards.
Since the book is at rest, these two forces are equal to each other and according to Newton's Third Law the reaction force to the force of gravity is equal but opposite to the weight of the book. This reaction force is the one that holds the book up on the shelf.
An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:
1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage. This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.
2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.
3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP. This newly made ATP can now be broken down to issue energy to fuel activity.
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
<h2>5.6kW</h2>
Explanation:
Step one:
given
mass m= 24kg
distance moved= 6m
time taken= 4seconds
Step two:
Required
power
but work done is the force applied at a distance, and the power is the work done time the time taken
Work done= F*D
F=mg
W= mg*D
W=24*9.81*6
W=1412.6J
Power P= work * time
P=1412.6*4
p=5650.5W
P=5.6kW
