A bicycle's performance, in both biological and mechanical terms, is extraordinarily efficient. ... 5 W/kg is about the level reachable by ordinary male athletes for longer ... This includes the power needed just for living, called the basal metabolic ... On a racing bicycle, a reasonably fit rider can ride at 40 km/h ( 25 mph) on flat ...
A motorcycle and 66.0 kg rider accelerate at 2.0 m/s2 up a ramp inclined 5.0° ... How much force is needed to accelerate a 50kg rider and his 250kg motorcycle at .... 5 months, she left the job and received $250 plus the motorcycle for her pay.
9.81 has three sig figs. .0050 is two figs, the 50 on the end .0050 * 9.981 = .04905 which to two sig figs is .049 N so 0.049 ...
How much force is needed to accelerate a 50-kg rider and her 250-kg motorcycle at 5 m/s2? This is really a physics question. The equation that you use is ...
So for our 2.0 cm driving wheel with an angular acceleration of 7.2 rad/s/s, and the 25.0 ... 65 revolutions as the car reduces its speed uniformly from 100km/h to 50km/h. .... This force is much greater at the locale of the bolt head itself, with a diameter of 15 ..... The pulley is a uniform cylinder of radius 0.26m and mass 7.50 kg.
Together, a motorcycle and its rider have a mass of 375 kg. If they are traveling north, what net force would be required to cause an acceleration of 4.50 m/s2, ...
Calculate the acceleration and the net force acting upon the car. ... Fnet = m•a = ( 500. kg)•(27.0 m/s2) = 13500 N (part b). Since the ... A 53.5-kg rider on a roller coaster car is moving 10.3 m/s at the top of a loop which has a radius of curvature of 7.29 m. ... Determine the minimum coefficient of friction required to keep a 920.
How much does the velocity change in 2 seconds? Answer: ... Question: Brakes are applied on a car of mass 1000 kg moving with a velocity of 54 km/h. The car covers a distance of 50 m before coming to rest. Calculate ... Question: What force would be needed to produce an acceleration of 4 m/s2 on a ball of mass 6 kg?
To check the no slip assumption, we calculate the maximum friction force fmax = µ N .... (14 kg)a. Solving, we have a = 2.01 m/s2. Since a > 0, our assumption is correct. ... can find the acceleration of the box (assumed to be down the incline). ..... cle and rider is 1601 N. The coefficient of kinetic fric ..... required to start slip is.