When the voltage across a steady resistance is doubled, the current?
1 answer:
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Answer:
The impact occured at a distance of 2478.585 meters from the person.
Explanation:
(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:
<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>
Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (), in seconds:
(1)
Where:
- Time spent by the sound in concrete, in seconds.
- Time spent by the sound in the air, in seconds.
By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:
(2)
Where:
- Speed of the sound in concrete, in meters per second.
- Speed of the sound in the air, in meters per second.
- Distance traveled by the sound, in meters.
If we know that ,
and
, then the distance travelled by the sound is:
The impact occured at a distance of 2478.585 meters from the person.
Answer:
i) The motion of both particles are shown on the same speed-time curve included
ii) Approximately 19.5 seconds
Explanation:
We are given that;
Initial velocity of particle, P = 140 m/s
Start time of particle P = 6 s before start time of particle Q
Position of particle Q when velocity is 25 m/s = 125 m
Therefore, from the equation of motion, we have for particle Q;
v² = u² + 2·a·s
Where:
v = Final velocity = 25 m/s
u = Initial velocity = 0 m/s
a = Acceleration
s = Distance covered = 125 m
Therefore;
25² = 0² + 2×a×125
Which gives a = 25²/(2×125) = 2.5 m/s²
The time taken for particle Q to reach 125 m is found from the relation;
s = u·t + 1/2·a·t²
Where:
t = Time of journey
Therefore;
125 = 0×t + 1/2×2.5×t²
Which gives 125 = 1.25 × t²
Hence, t² = 125/1.25 = 100
t = √(100) = 10 s
The equation for particle Q is v = 0 + 2.5×t
Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;
Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s
From the equation of motion, for particle P (decelerating) we have
v = u - a·t
Where:
v = 25 m/s
u = 140 m/s
t = 16 s
Hence, 25 = 140 - a×16
∴ 16·a = 140 - 25 = 115
a = 115/16 = 7.1875 m/s²
Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;
v = u - a·t
0 = 140 - 7.1875 × t
∴7.1875·t = 140
t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.
