A boat has a mass of 6800 kg?
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
602447
6007.6 using 1 kg = 2.20462262185 lbs
588958
the weight will be according 2 density of the material \n\nfor example for water 1  liter wighs 1 kg
325473
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)
351568
6.4185×1023 kg, or the equivalent of 0.107 Earths.
985921
It looks kind of like a boat. See the source.
257654
Say the anchor weighs 15 pounds in air and displaces 2 pounds of water.\nBefore the drop, the boat displaces its weight plus the 15 lb weight in air of the anchor. This is the maximum displacement situation, so the water is at maximum height.\nAfter the drop, if the anchor hits bottom, the boat displaces its weight only and the anchor displaces 2 lb of water, the minimum displacement situation, so the water is at minimum height.\nAfter the drop, if the anchor doesn't reach the bottom, the boat displaces its weight plus the 13 lb weight in water of the anchor. The anchor displaces 2 lb water. The total displacement is the same as the maximum displacement situation, so the water is at maximum height.
666681
1/2 * mass * velocity^2.
762043
Use the conservation of momentum:\n\nM * (change in velocity) + (change in mass) * V = 0\n\nM * dv + dm * V = 0\n\nWhere M is the mass of the child and boat system, dv is the boat's change in velocity, dm is the change in mass of the initial system (the package) and V is the speed at which the package is ejected.\n\nSubstituting the values:\n\n(26 kg + 45 kg) * V = - (6.4 kg) * (10 m/s)\n\n(71 kg) * V = - (64 kg-m/s)\n\nSolving for V:\n\nV = - (64 kg-m/s) / (71 kg)\n\nV = -.9014 m/s\n\nThus, the boat travels backwards at a speed of .9014 m/s. That's it! :)
293494
A big boat that carries oil.
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
120890
About 6,000 pounds per square inch
244521
there is an engine i saw that has no moving parts. just magnets that attract water in one end and out the other, but it only works in salt water.
1078409
Someone with the ability to make & pilot a boat. :P
871908
grumman is a good boat
947129
a) Force on the box due to max acceleration witout moving the box = the maximum frictional force = 0.24* Normal force = 0.24*Weightof box = 0.24*mg\n\nSo max acceleration = 0.24*g = 2.352m/s^2\ns = .5at^2 = 10.584\n\nb)\nthis is a vector problem\nthe boat needs to be driven towards west inorder to cross exactly to a point at north.\nresultant speed of boat = 140/20 = 7m/s\nso using pythagorian theorem, U^2 = 7^2 +4^2, or U=sqrt(65) = 8.0623\n\nc)displacement = (11i - 5j) - (7i - 8j)  = (4i + 3j) \nforce = (3i + 4j)\nwork done = Fs = (3i + 4j)(4i + 3j)  = 12 + 12=24\n\nSo change in Kinectic energy  = work done  = 24J
473092
boat Specifications: \nYear 2002 \nLength Overall (Feet) 22 \nDisplacement 3150 \nFuel Capacity 45 \nSeating Capacity 11 \nAcceleration to 36 MPH (seconds) 6.6 seconds; 189.4 feet \nNoise at 36 MPH 86 D 86 R \nNoise at Idle 58 D 59 R \nTest Engine PCM Excalibur EFI 330hp \nTest Prop 14 x 18 OJ 4-blade \nTop Speed 44.2 \n \nboat Description: \nThis is MB’s top-of-the-line wakeboarding boat for core wakeboarders. According to the manufacturer, this is a pure performance solution for wakeboarders with one goal in mind – bigger wakes.