A 2200 kg car is moving due East at 25 m/s when the driver applies the brakes to

Question

A 2200 kg car is moving due East at 25 m/s when the driver applies the brakes to slow down to 15 m/s. If the car moves a distance to the Eastward of 100 m during this slowing process, what was the magnitude of the average net slowing force on the car during the time the brakes were applied? Several forces act on an object at the instant it is moving with a velocity of v = -6i + 2 j m/s. One of the forces that acts is the constant force F = 4i - 3j N. Please calculate the angle between the direction of the velocity and the force F vector at this moment, The intensity of sunlight falling on the ground in Tennessee on a nice bright summer day is 1100 W/m^2. A solar panel at your farm is an amazing 22% efficient-that is, in manages to convert 22% of the energy that falls on it into electricity. One of your solar panels is a huge square 10 m by 10 m. Calculate the maximum power output of your panel on a nice, cloudless Tennessee day when the sunlight falls straight down on your panel.

Explanation / Answer

work done = F*x

from work energy theorem

work done = change in KE

F*x = 0.5*m*(v2^2-v1^2)

F*100 = 0.5*2200*(15^2-25^2)

F = 4400 N <<--------answer

option (d)

+++++++++++++++++++

magnitude of velocity lvl = sqrt(6^2+2^2) = 6.32 m/s

magnitude of force lFl = sqrt(4^2+3^2) = 5 m/s

v.F = (-6i+2j) . (4i - 3j)

v.F = (-24-6) = -30

angle = cos^-1(v.F/lvl*lFl)

direction = 162 degrees

9)

power absorbed = I*A = I*l^2 = 1100*10^2 = 11*10^4 W

maximum power out put = 0.22*Pmax = 0.22*11*10^4 = 24200 W = 24.2 kW

answer (d)

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