The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.
For this "type" of motion, displacement (Δx) can be determined by:
is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
= - 42
Displacement of the boat for t=6.0s interval is = - 42m, i.e., 42 m to the left.
Answer:
x = 1,185 m
, t = 4/3 s
, F = - 4 N
Explanation:
For this exercise we use Newton's second law
F = m a = m dv /dt
β - α t = m dv / dt
dv = (β – α t) dt
We integrate
v = β t - ½ α t²
We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t
v-v₀ = β t - ½ α t²
the farthest point of the body is when v = v₀ = 0
0 = β t - ½ α t²
t = 2 β / α
t = 2 4/6
t = 4/3 s
Let's find the distance at this time
v = dx / dt
dx / dt = v₀ + β t - ½ α t2
dx = (v₀ + β t - ½ α t2) dt
We integrate
x = v₀ t + ½ β t - ½ 1/3 α t³
x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³
The body comes out of rest
x = 3.5556 - 2.37
x = 1,185 m
The value of force is
F = β - α t
F = 4 - 6 4/3
F = - 4 N
Wow ! This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.
The centripetal acceleration of any object moving in a circle is
(speed-squared) / (radius of the circle) .
Notice that we won't need to use the mass of the train.
We know the radius of the track. We don't know the trains speed yet,
but we do have enough information to figure it out. That's what we
need to do first.
Speed = (distance traveled) / (time to travel the distance).
Distance = 10 laps of the track. Well how far is that ? ? ?
1 lap = circumference of the track = (2π) x (radius) = 2.4π meters
10 laps = 24π meters.
Time = 1 minute 20 seconds = 80 seconds
The trains speed is (distance) / (time)
= (24π meters) / (80 seconds)
= 0.3 π meters/second .
NOW ... finally, we're ready to find the centripetal acceleration.
<span> (speed)² / (radius)
= (0.3π m/s)² / (1.2 meters)
= (0.09π m²/s²) / (1.2 meters)
= (0.09π / 1.2) m/s²
= 0.236 m/s² . (rounded)
If there's another part of the problem that wants you to find
the centripetal FORCE ...
Well, Force = (mass) · (acceleration) .
We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force. </span>
Answer:
Explanation:
Let t be the final temp
temp. drop of nickel=147.4-t
temp. gain by water=t-29.9
in equilibrium
Heat lost=Heat gained
58.7×0.443×(147.4-t)=184×4.18×(t-29.9)
58.7×0.443×147.4-58.7×0.443×t=184×4.18×t-184×4.18×29.9
3833.00434 -26.0041 t=769.12t-22996.688
(769.12 +26.0041)t= ?
find t
