------------ 2) The kinetic energy K (in J) of an object is given by K=1/2mv^2,
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Question
------------2) The kinetic energy K (in J) of an object is given by K=1/2mv^2, where m is the mass (in kg) of the object and v is its velocity. If a 250-kg wrecking ball accelerates at 5.0m/s^2, how fast is the kinetic energy changing when v=30.0 m/s? ----------8) An appro. relationship between the pressure p and volume v of the vapor in a diesel engine cylinder is pv^1.4 = k, where k is a constant. At a certaain instant, p=4200 kPa, v=75 cm^3, and the vol is increasing at the rate of 850cm^3 /s. What is the time rate of change of the pressure at this instant?
------------10) A computer program increases the side of a sq. image on the screen at the rate of 0.25 in/s. Find the rate at which the area of the image increases when the edge is 6.50 in.
>Need a step by step solution. Thank you.
------------2) The kinetic energy K (in J) of an object is given by K=1/2mv^2, where m is the mass (in kg) of the object and v is its velocity. If a 250-kg wrecking ball accelerates at 5.0m/s^2, how fast is the kinetic energy changing when v=30.0 m/s? ----------8) An appro. relationship between the pressure p and volume v of the vapor in a diesel engine cylinder is pv^1.4 = k, where k is a constant. At a certaain instant, p=4200 kPa, v=75 cm^3, and the vol is increasing at the rate of 850cm^3 /s. What is the time rate of change of the pressure at this instant?
------------10) A computer program increases the side of a sq. image on the screen at the rate of 0.25 in/s. Find the rate at which the area of the image increases when the edge is 6.50 in.
>Need a step by step solution. Thank you.
Explanation / Answer
2) K = 1/2 m v^2 dK/dt = mv*dv/dt but dv/dt = acceleration = 5 m /s^2 and given v = 30m/s and m= 250kg hence dK/dt = rate of change of kinetic energy = 250*30*5 = 37.5 KW 8) pv^1.4 = k at p = 4200kpa= 4200*10^3 pa and v = 75 cc = 75*10^(-6) m^3 k = 7.05 hence pv^1.4 = 7.05 v^1.4 = 7.05 /p 1.4v^0.4 * dv = (-7.05/p^2 )*dp v = 850cc = 850*10^(-6) m^3 hence dp = - 66662.4 kpa 10) A = s^2 dA/dt = 2s*ds/dt s = 6.5 in and ds/dt = 0.25in/s dA/dt = 2*6.5*0.25 = 3.25in^2 /s
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