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37. A skateboarder (m = 85 kg) takes a running jump onto a friend's skateboard that is initially at rest as sketched in Figure P7.37. The friend is standing on the skateboard (mass of friend plus skateboard 110 kg). After landing on the skateboard, the velocity of the board plus the two skateboarders is 3.0 m/s. What was the norizontal component of the velocity of the jumping skateboarder just before he landed on the skateboard? ndw 0.1 alb bosqdw .noi %3D Figure P7.37 of 300 m/s

37. A skateboarder (m = 85 kg) takes a running jump onto a friend's skateboard that is initially at rest as sketched in Figure P7.37. The friend is standing on the skateboard (mass of friend plus skateboard 110 kg). After landing on the skateboard, the velocity of the board plus the two skateboarders is 3.0 m/s. What was the norizontal component of the velocity of the jumping skateboarder just before he landed on the skateboard? ndw 0.1 alb bosqdw .noi %3D Figure P7.37 of 300 m/s

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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#37. Attached. Must show diagram FBD and all math and formula manipulation. Must use equation starting off this sheet attached. And then can manipulate it as you wish. Must start from equation sheet.
velocity of 300 m/s at an angle of Find the recoil velocity of the car. speed of 14 m/s on a frictionless, horizontal track. It is raining, kg with a and water begins to accumulate in the car. After some time, it is found that the speed of the car is only 11 m/s. How much water 35. An open railroad car of mass 2500 kg is coasting with an initial 36. * The M79 grenade launcher was first used by the U.S. Army in 40° with respect to the horizontal. Figure P7.34 42 (in kilograms) has accumulated in the car? 1961. It fires a grenade with an approximate mass m 7.0 kg at a speed of 75 m/s. (a) If a soldier fires the M79 horizontally while standing on a very slippery surface, what is the recoil speed of rhe soldier? Take the mass of the soldier plus M79 to be 100 kg. (b) The barrel of the M79 is 36 cm long. Estimate the average recoil force on the soldier. 37. A skateboarder = 85 kg) takes a (m : running jump onto a friend's skateboard that is initially at rest as sketched in Figure P7.37. The friend is standing on the skateboard (mass of friend plus skateboard: 110 kg). After landing on the skateboard, the velocity of the board plus the two skateboarders is 3.0 m/s. What was the horizontal component of the velocity of the jumping skateboarder just before he landed on the skateboard? 38. Figure P7.37 A bullet of mass 15 g is fired with an initial speed of 300 m/s into a wooden block that is initially at rest. The bullet becomes lodged in the block, and the bullet and block then slide together on the floor for a distance of 1.5 m before coming to rest (Fig. P7.38). If the coefficient of friction between the block and the floor is 0 40 what is the mass of the block?
Transcribed Image Text:velocity of 300 m/s at an angle of Find the recoil velocity of the car. speed of 14 m/s on a frictionless, horizontal track. It is raining, kg with a and water begins to accumulate in the car. After some time, it is found that the speed of the car is only 11 m/s. How much water 35. An open railroad car of mass 2500 kg is coasting with an initial 36. * The M79 grenade launcher was first used by the U.S. Army in 40° with respect to the horizontal. Figure P7.34 42 (in kilograms) has accumulated in the car? 1961. It fires a grenade with an approximate mass m 7.0 kg at a speed of 75 m/s. (a) If a soldier fires the M79 horizontally while standing on a very slippery surface, what is the recoil speed of rhe soldier? Take the mass of the soldier plus M79 to be 100 kg. (b) The barrel of the M79 is 36 cm long. Estimate the average recoil force on the soldier. 37. A skateboarder = 85 kg) takes a (m : running jump onto a friend's skateboard that is initially at rest as sketched in Figure P7.37. The friend is standing on the skateboard (mass of friend plus skateboard: 110 kg). After landing on the skateboard, the velocity of the board plus the two skateboarders is 3.0 m/s. What was the horizontal component of the velocity of the jumping skateboarder just before he landed on the skateboard? 38. Figure P7.37 A bullet of mass 15 g is fired with an initial speed of 300 m/s into a wooden block that is initially at rest. The bullet becomes lodged in the block, and the bullet and block then slide together on the floor for a distance of 1.5 m before coming to rest (Fig. P7.38). If the coefficient of friction between the block and the floor is 0 40 what is the mass of the block?
Must start wlone of these eguahons Por each poblen and shou all addihons, manipulahou of forrula or esery step and all alsebra. Physics 103 Equation Sheet G=6.67 x 1011 Nm?/kg? Patm= 1.01 x 10$ N/m2 MEarth= 5.98 x 1024 kg g = 9.8 m/s? = 32 ft/s? k = 1.38 x 1023 J/K O = 5.67 x 10 W/m²K* Pwater = 1.0 x 10³ kg/m³ Pair = 1.29 kg/m Pseawater = 1.03 x 10 kg/m Pice = 0.92 x 10³ kg/m³ Piron = 7.87 x 10³ kg/m³ Paluminum = 2.7 x 10 kg/m3 sin e = opp/hyp a? + b? = c? cos 0 = adj/hyp tan 0 = opp/adj -b t Vb2 – 4ac 2a Ax Av a = At Δω a = At Δt v = vo? + 2a(x-Xo) Δω At x = Xo + Vot + ½ at? V = Vo + at EF = ma Στ- Ια T= Fr sin 0 F = w = mg Gm,m2 F, = Frr S uN ac = v/r Fb = pVsubg Fsp = -kAx r2 P = Po+ pgh P = F/A p = m/V VA1 = V2A2=AV/At Pi+pghi+% pv;?= Pz+pgh;+% ov? Vave = (AP r*)/8ŋL P = W/At FAt = Ap W = FAr cos 0 Wnet = AKE KE, + PE1 = KE2+ PE2 - Wnc PE = mgh KE = % mv? PESP = % k Ax? p= mv E = mc? mOmmentum. T2-T1 = KA L. = GEA(T – T) AS = Q = mc AT KEave = 3/2 kT Q = ml AU = Q-W W = PAV R=L/K PV = NkT thermeff = 1- Tc/TH WAh one above Fnet mH=MAV/AT Fnet AT= AMV AP - AMV Fnet AT
Transcribed Image Text:Must start wlone of these eguahons Por each poblen and shou all addihons, manipulahou of forrula or esery step and all alsebra. Physics 103 Equation Sheet G=6.67 x 1011 Nm?/kg? Patm= 1.01 x 10$ N/m2 MEarth= 5.98 x 1024 kg g = 9.8 m/s? = 32 ft/s? k = 1.38 x 1023 J/K O = 5.67 x 10 W/m²K* Pwater = 1.0 x 10³ kg/m³ Pair = 1.29 kg/m Pseawater = 1.03 x 10 kg/m Pice = 0.92 x 10³ kg/m³ Piron = 7.87 x 10³ kg/m³ Paluminum = 2.7 x 10 kg/m3 sin e = opp/hyp a? + b? = c? cos 0 = adj/hyp tan 0 = opp/adj -b t Vb2 – 4ac 2a Ax Av a = At Δω a = At Δt v = vo? + 2a(x-Xo) Δω At x = Xo + Vot + ½ at? V = Vo + at EF = ma Στ- Ια T= Fr sin 0 F = w = mg Gm,m2 F, = Frr S uN ac = v/r Fb = pVsubg Fsp = -kAx r2 P = Po+ pgh P = F/A p = m/V VA1 = V2A2=AV/At Pi+pghi+% pv;?= Pz+pgh;+% ov? Vave = (AP r*)/8ŋL P = W/At FAt = Ap W = FAr cos 0 Wnet = AKE KE, + PE1 = KE2+ PE2 - Wnc PE = mgh KE = % mv? PESP = % k Ax? p= mv E = mc? mOmmentum. T2-T1 = KA L. = GEA(T – T) AS = Q = mc AT KEave = 3/2 kT Q = ml AU = Q-W W = PAV R=L/K PV = NkT thermeff = 1- Tc/TH WAh one above Fnet mH=MAV/AT Fnet AT= AMV AP - AMV Fnet AT
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