Answer:
Explanation:
Given parameters:
Weight of object = 49N
Force applied = 12N
Unknown:
Acceleration of object = ?
Solution:
The acceleration of the object is found by dividing the force by the weight;
Acceleration = [tex]\frac{12}{49}[/tex] = 0.25m/s²
Help please!!!
if superman at 90kg jumps a 40m building in a single bound how much work does superman perform
Answer:
5 years worth of work (aka all of the homework i currently have)
Electric power is to be generated by installing a hydraulic turbine-generator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily. If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.
Answer:
[tex]\eta_{turbine} = 0.777 = 77.7\%[/tex]
[tex]\eta_{combined} = 0.728 = 72.8\%[/tex]
Explanation:
First we calculate the power input to the turbine. The input power will be equal to the potential energy of water per unit time:
Input Power = [tex]P_{in} = \frac{Work}{Time} = \frac{Potential\ Energy\ of\ Water}{t} \\P_{in} = \frac{(mass)(g)(height)}{Time} = (mass flow rate)(g)(height)\\\\P_{in} = (1500\ kg/s)(9.81\ m/s^2)(70\ m)\\P_{in} = 1.03\ x\ 10^6\ W = 1030\ KW[/tex]
Now, for turbine efficiency:
[tex]\eta_{turbine} = \frac{Mechanical\ Power\ Out}{P_{in}}\\\\\eta_{turbine} = \frac{800\ KW}{1030\ KW}\\\\\eta_{turbine} = 0.777 = 77.7\%[/tex]
for generator efficiency:
[tex]\eta_{generator} = \frac{Power\ Generation}{Mechanical\ Power\ Out}\\\\ \eta_{generator} = \frac{750\ KW}{800\ KW}\\\\\eta_{turbine} = 0.9375 = 93.75\%[/tex]
Now, for combined efficiency:
[tex]\eta_{combined} = \eta_{turbine}\ \eta_{generator}\\\\\eta_{combined} = (0.777)(0.937)\\\eta_{combined} = 0.728 = 72.8\%[/tex]
what power is transmitted by 2A flowing across 5V
Answer:
[tex]\huge\boxed{10\:watts}[/tex]
Explanation:
Power = Current × Voltage
P = I × V
P = 2 × 5
P = 10 watts
