#3 = KE = 1/2 mv^2 + 1/2 Iw^2 = 1/2 (7.0 kg)(32.5 m/s)^2 + 1/2 (2.8 x 10^-2 kg·m^2)(32.5 m/s/0.10 m)^2 = 837.5 J
#4 = The maximum number of buses the cyclist can jump over is equal to the horizontal distance the cyclist can travel divided by the width of each bus. The horizontal distance the cyclist can travel is equal to the initial velocity multiplied by the time it takes for the cyclist to reach the landing ramp. The time it takes for the cyclist to reach the landing ramp is equal to the vertical distance divided by the initial velocity multiplied by the sine of the angle of the ramp. Therefore, the maximum number of buses the cyclist can jump over is equal to (32.5 m/s)((2.74 m/18.8°) / (32.5 m/s)sin(18.8°)) = 8.3 buses.
1.What is the gravitational force between two trucks, each with a mass of 2.0 x 10^4 kg, that are 2.0 m apart? (G = 6.673 x 10^11 N•m^2/kg^2)
2.Each tire of an automobile has an area of 0.026 m^2 in contact with the ground. The weight of the automobile is 2.6 10^4 N. What is the pressure in the tires?
3.A bowling ball has a mass of 7.0 kg, a moment of inertia of 2.8 × 10^-2 kg·m^2 and a radius of
0.10 m. If it rolls down the lane without slipping at a linear speed of 4.0 m/s, what is its
total kinetic energy?
1. Use Newtons Gravitaional law.
2. Pressure = weight on tire/ area
3. KE = 1/2 mv^2 + 1/2 Iw^2
but w= v/r
so KE= ......
what is the gravitational force between two trucks, each with a mass of 2.0*10^4kg, that are 2.0 m apart?
a motorcycle daredevil is attempting to jump across as many buses as possible. The take off ramp makes an angle18.8dreg above the horizontal, and the landing rampis identical t the take off ramp. The buses are park side by side, and each bus is 2.74m wide. The cyclist leaves the ramp with a speed of v=32.5m/s. What is the maximum number of buses over which the cyclist can jump?
#2 = 250000Pascal
1 answer
1 answer