In a downhill ski race surprisingly little advantage is gained by getting a running start. This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills. To demonstrate this, find the final speed and the time taken for a skier who skies 75.0 m along a 35° slope neglecting friction for the following two cases. (Note that this time difference can be very significant in competitive events so it is still worthwhile to get a running start.) (a) starting from rest final speed m/s time taken s (b) starting with an initial speed of 3.00 m/s final speed m/s time taken s

In a downhill ski race surprisingly little advantage is gained by getting a running start. This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills. To demonstrate this, find the final speed and the time taken for a skier who skies 75.0 m along a 35° slope neglecting friction for the following two cases. (Note that this time difference can be very significant in competitive events so it is still worthwhile to get a running start.) (a) starting from rest final speed m/s time taken s (b) starting with an initial speed of 3.00 m/s final speed m/s time taken s

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In a downhill ski race surprisingly little advantage is gained by getting a running start. This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills. To demonstrate this, find the final speed and the time taken for a skier who skies 60.0 m along a 25° slope neglecting friction for the following two cases. (Note that this time difference can be very significant in competitive events so it is still worthwhile to get a running start.) (a) starting from rest_____ final speed m/s_____ time taken s(b) starting with an initial speed of 2.50 m/s_____ final speed m/s_____ time taken s
In a downhill ski race, little advantage is gained by getting a "running" start. This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills. To demonstrate this, find the final speed of and the time taken for a skier who skis 70 m along a 30° slope neglecting friction: (a) starting from rest v = __ m/s t= __ s (b) starting with an initial speed of 2.6 m/s, compute the ratio of the gain in |ΔPE| when the skier reaches the bottom of the hill to the initial KE initial. |ΔPE|KEinitial= (c) starting with an initial speed of 2.6 m/s v= __ m/s t= __ s
A 68.1 kg runner has a speed of 3.50 m/s at one instant during a long-distance event. For (a) Apply the definition of kinetic energy. For (b) use the work-energy theorem. (a) What is the runner's kinetic energy at this instant (in J)? (b) How much net work (in J) is required to triple her speed?
please answer this question according to the blanks: A falling package with a parachute is greatly affected by air resistance. Suppose a package (m = 25 kg), dropped from an altitude of y = 1500 m, hits the ground at a speed of v = 45 m/s. Calculate the work done by air resistance. By conservation of energy, the initial total mechanical energy is equal to the final total mechanical energy. Let K be the kinetic energy, U be the gravitational potential energy, and Wair be the work done by the drag force from air resistance. E1 = E2 K1 + U1 + Wair = K2 + U2 Some of the terms in the equation above are zero so it can be simplified to: Blank 1 + Wair = Blank 2 Isolating the work done by air resistance, we get: Wair = ½Blank 3 - mBlank 4 Plugging in the values given, the work done by air resistance is: Wair = Blank 52.19 kJ
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A falling package with a parachute is greatly affected by air resistance. Suppose a package (m = 25 kg), dropped from an altitude of y = 1500 m, hits the ground at a speed of v = 45 m/s. Calculate the work done by air resistance. By conservation of energy, the initial total mechanical energy is equal to the final total mechanical energy. Let K be the kinetic energy, U be the gravitational potential energy, and Wair be the work done by the drag force from air resistance. E1 = E2 K1 + U1 + Wair = K2 + U2 Some of the terms in the equation above are zero so it can be simplified to: + Wair Isolating the work done by air resistance, we get: Wair = 2 - m Plugging in the values given, the work done by air resistance is: Wair - 34 2.19 k)
On Earth with a gravational acceleration g, the potential energy stored in an object varies directly with its mass m and its vertical height h. What is the equations of the potential energy of a 2kg skateboard that is sliding down a ramp?
A falling package with a parachute is greatly affected by air resistance. Suppose a package (m = 25 kg), dropped from an altitude of y = 1500 m, hits the ground at a speed of v = 45 m/s. Calculate the work done by air resistance. By conservation of energy, the initial total mechanical energy is equal to the final total mechanical energy. Let K be the kinetic energy, U be the gravitational potential energy, and Wair be the work done by the drag force from air resistance. E1 = E2 K1 + U1 + Wair = K2 + U2 Some of the terms in the equation above are zero so it can be simplified to: _____ + Wair = _____ Isolating the work done by air resistance, we get: Wair = ½_____ - m _____ Plugging in the values given, the work done by air resistance is: Wair =_____ 2.19 kJ