I hope it is clearly visible.. Velocity of the center of mass of 2-ball system is - 11.54m/s. Minus indicates, velocity direction is in downward direction.
Let F = the downstream speed of the water.
<span>Then the boat's upstream speed is: 15 - F </span>
<span>The boat's downstream speed is: 15 + F </span>
<span>Assume both the journeys mentioned take T hours, then using "speed x time = distance" we get: </span>
<span>Downstream journey: (15 + F)T = 140 </span>
<span>Upstream journey: (15 - F)T = 35 </span>
<span>Add the two formulae together: </span>
<span>(15 + F)T + (15 - F)T = 140 + 35 </span>
<span>15T + FT + 15T - FT = 175 </span>
<span>30T = 175 </span>
<span>T = 35/6 </span>
<span>Use one of the equations to find F: </span>
<span>(15 + F)T = 140 </span>
<span>15 + F = 140/T </span>
<span>F = 140/T - 15 </span>
<span>F = 140/(35/6) - 15 </span>
<span>F = 24 - 15 </span>
<span>F = 9 </span>
<span>i.e. the downstream speed of the water is 9 kph </span>
<span>Therefore, the boat's speed downstream is 15 + F = 15 + 9 = 24 kph.
the answer is: *24kph*</span>
Answer:
See bolded below.
Explanation:
Consider the " Before " and " After. " " Before, " this particle 1 was trying to catch up with this particle 2, and " after " particle one had collided with particle two. Take a look at the attachment below for a more detailed examination.
Here is how this will play out. Particle 1, with great velocity, will hit particle 2, which would mean that Particle 2 has less velocity than Particle 1. Now after the collision, energy is transferred to Particle 2, and while Particle 1 has now stopped in it's tracks, Particle 2 - with more energy than before - will continue as long as it has to before friction eventually brings it to a stop.
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From this we can conclude that Vf, from the picture below, must have less energy than V1, but more energy than V2 - and vice versa.
Answer:
Different forces (including magnetism, gravity, and friction) can affect motion
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s