what is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?
1 answer:
The initial velocity of go-kart is 2.5 m/s.
<u>Explanation:</u>
Here, the uniform acceleration of go-kart is given as 0.5 m/s². Also the time required by it to stop is also given as 5 s. As acceleration is the measure of change in velocity per unit time.
In this case, the velocity should be changed from a value to zero to come to rest. So the initial velocity will be positive value and final velocity is zero.
As we know the values of acceleration, final velocity and time, the initial velocity can be easily determined as follows.
Since, final velocity is zero, acceleration is 0.5 m/s² and time is 5 s, then,
Initial velocity = 0.5 × 5 = 2.5 m/s.
So the initial velocity of go-kart is 2.5 m/s.
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Answer:
(a) = 0.22 W
(b) = 0.056 W
Explanation:
given information:
the mass of piano wire, m = 3.00 g = 0.003 kg
tension, F = 25 N
length, l = 80 cm = 0.8 m
frequency, f = 120 Hz
amplitude, A = 1.6 mm = 0.0016 m
(a) the average power carried by the wave,
=
(√μF)ω²A²
where,
ω = 2πf = 2π120 = 754
μ =
=
= 0.00375 kg/m
thus,
=
(√(0.00375)(25))(754)²(0.0016)²
= 0.22 W
(b) What happens to the average power if the wave amplitude is halved.
based on the equation above, we know that the average power is proportional to the square amplitude. therefore
=
= 0.056 W
(a) The initial speed at which the bottles are launched is 4.27 m/s.
(b) The horizontal displacement at which the bottle land is 1.75 m.
<h3>Initial speed of the bottle</h3>The initial speed of the bottle is calculated as follows;
where;
- T is time of flight
- u is the initial speed
2usinθ = Tg
u = Tg/(2sinθ)
u = (0.5 x 9.8)/(2 x sin35)
u = 4.27 m/s
<h3>Horizontal displacement of the bottle</h3>X = u²sin(2θ)/g
X = (4.27² x sin(70))/(9.8)
X = 1.75 m
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