Downhill

Lehman, M.D., M.S.P.H. Participantes do Projeto Piloto Xavier Amador, Ph.D. Nancy Coover Andreasen, M.D., Ph.D. F. M. Baker, M.D. Donald W. Black, M.D. Carlos S. Castillo, M.D. Scott C. Clark, M.D. William Coryell, M.D. Lisa B. Dixon, M.D. Jack E. Downhill, Jr., M.D. Katherine P. Duffy, M.D. Jean Endicott, Ph.D. Michael A. Fauman, M.D., Ph.D. Miriam Gibbon, M.S.W. Jack Gorman, M.D. Paul E. Hogsten, M.D. Michael L. Jeffries, M.D. Douglas Langbehn M.D. Joseph Liberto, M.D. David B. Mallot,…

gravity does work at the rate of Pg = 8.00 hp = 5968 W Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the…

does work at the rate of Pg = 8.00 hp = 5968 W 6.100. Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in…

does work at the rate of Pg = 8.00 hp = 5968 W 6.100. Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in…

does work at the rate of Pg = 8.00 hp = 5968 W 6.100. Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in…

does work at the rate of Pg = 8.00 hp = 5968 W 6.100. Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in…

does work at the rate of Pg = 8.00 hp = 5968 W 6.100. Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in…

does work at the rate of Pg = 8.00 hp = 5968 W 6.100. Pg 5968 W = = 0.02030 mgv (1800 kg)(9.80 m/s 2 )(16.67 m/s) α = 1.163° and tan α = 0.0203, a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is mg sin α = 358 N and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in…