The answer is Newton's 3rd Law. The reason why is because a force is a push or a pull that acts upon an object as a results of its interaction with another object. ... These two forces are called action and reaction forces and are the subject ofNewton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction.
Answer:
option (C)
Explanation:
The amount of heat required to raise the temperature of substance of unit mass by unit degree is called specific heat of that substance.
Its SI unit is Joule / Kg °C.
Every material has a constant value of specific heat.
So, option (c) is correct.
When an object absorbs an amount of energy equal to Q, its temperature raises by
following the formula
where m is the mass of the object and
is the specific heat capacity of the material.
In our problem, we have
,
and
, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
Learn more about current:
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Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
