Answer:
68.25 kJ
Explanation:
The thermal energy Q required to raise the temperature of 15kg gold from 45⁰ C up to 80⁰ C is Q = mcΔθ where m = mass of gold = 15 kg, c = specific heat capacity of gold = 130 J/kg°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = 45 °C and θ₂ = 80 °C
So, Q = mcΔθ = mc(θ₂ - θ₁)
= 15 kg × 130 J/kg°C × (80 °C - 45 °C)
= 1950 J/°C × 35 °C
= 68,250 J
= 68.25 kJ
The range of force exerted at the end of the rope is; 285.7 N to 1,000 N.
<h3>What is the Net horizontal force?</h3>
The net horizontal force of the cylinder when it is at equilibrium position can be found by applying Newton's second law of motion. Thus;
∑F = 0
F - μF_n = 0
We are given;
F_n = 5 kN = 5000 N
μ = 0.2
Thus;
F - 0.2(5,000) = 0
F - 1,000 = 0
F = 1,000 N
The strength of the applied force will be increasing as the number of turns of the rope increases. Thus;
Minimum force = Total force/number of turns of rope
Since rope is wrapped three and half times, then;
number of turns = 3.5
Thus;
minimum force = 1,000/3.5
minimum force = 285.7 N
Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.
Read more about net horizontal force at; brainly.com/question/26957287
Answer: Time to fill tank= 181.983s
Mass flow rate of gasoline = 0.1556kg/s
Explanation:
Answer:
Explanation:
The image that is supposed to be attached to the question is displayed in the diagram below.
Applying Nodal Analysis at node 1;
where;
(from the circuit)
= 
= ![V_o [ \dfrac{1}{12.5}+\dfrac{1}{50}+\dfrac{1}{10}-\dfrac{75}{500}] = \dfrac{50}{12.5}](https://tex.z-dn.net/?f=V_o%20%5B%20%5Cdfrac%7B1%7D%7B12.5%7D%2B%5Cdfrac%7B1%7D%7B50%7D%2B%5Cdfrac%7B1%7D%7B10%7D-%5Cdfrac%7B75%7D%7B500%7D%5D%20%3D%20%5Cdfrac%7B50%7D%7B12.5%7D)
= ![V_o[ \dfrac{500*500+12.5*5000+12.5*5000*5-75*12.5*500}{12.5*50*10*500}]= \dfrac{50}{12.5}](https://tex.z-dn.net/?f=V_o%5B%20%5Cdfrac%7B500%2A500%2B12.5%2A5000%2B12.5%2A5000%2A5-75%2A12.5%2A500%7D%7B12.5%2A50%2A10%2A500%7D%5D%3D%20%5Cdfrac%7B50%7D%7B12.5%7D)
= 
