Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
We know, the ideal gas equation,
P1V1 / T1 = P2V2 / T2
Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K
P2 = 400 mm Hg
T2 = -23 + 273 = 250 K
Substitute their values,
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83
In short, Your Answer would be approx. 15.83 m3
Hope this helps!
We Know, K.E. = 1/2 × m × v²
From the expression, we can conclude that Kinetic energy is directly proportional to mass. So, as mass will increase, Kinetic energy will also increase.
In short, Your Correct answer would be Option B
Hope this helps!
