A rocket burns fuel at a rate of 199 kg/s and exhausts the gas at a relative speed of 9 km/s. Find the thrust of the rocket. Ans
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Answer:
which corresponds to the second option shown: "voltage times amperage"
Explanation:
The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.
Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:
Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.
This corresponds to the second option shown in the question: "Voltage times amperage".
To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)
PART A) By definition the frequency in a spring is given by the equation
Where,
m = mass
k = spring constant
Our values are,
k=1700N/m
m=5.3 kg
Replacing,
PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.
Where,
k = Spring constant
m = mass
y = Vertical compression
v = Velocity
This expression is equivalent to,
Our values are given as,
k=1700 N/m
V=1.70 m/s
y=0.045m
m=5.3 kg
Replacing we have,
Solving for A,
PART C) Finally, the total mechanical energy is given by the equation