Answer: E = 941738.537J
Explanation:
to begin,
given that the mass = 2320 pound = 1052.334 kg
Δh = 110 ft = 33.528 m
given that distance (d) = 1283 ft = 391.058 m
also the speed (v) is 65 mph = 29.058 m/s
force (F) = 87 pounds = 386.995 N
from our knowledge in work energy theory;
E = Fd + 1/2mv² + mgh
E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)
E = 151337.491 + 444278.2 + 346122.84
E = 941738.537J
i hope this helps, cheers.
The sound wave will have traveled 2565 m farther in water than in air.
Answer:
Explanation:
It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.
Distance = Velocity × Time.
So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.
As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.
Distance = 343 × 2.25 =771.75 m
And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.
Distance = 1483×2.25=3337 m.
Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.
Difference in distance covered in water and air = 3337-772 m = 2565 m
So the sound wave will have traveled 2565 m farther in water than in air.
