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.
Assuming that all the bullet’s energy heats the paraffin, its final temperature is 27.1 degree C. The correct option is D.
<h3>What is temperature?</h3>
Temperature is the degree of hotness and coldness of the material.
The energy of the bullet E = 1/2 mv²
E = 1/2 x 10 x 10⁻³ x (2000)²
E = 2 x 10⁴ J
This heat is used in heating the paraffin
E = m x c ΔT = m x c (Tfinal -Tinitial)
2 x 10⁴ J = 1 x 2.8 x 10³ x (Tfinal -20)
Tfinal = 27.1°C
Thus, the final temperature is 27.1 degree C. The correct option is D.
Learn more about temperature.
brainly.com/question/15267055
#SPJ1
Answer:
if there is no friction in a simple machine, work output and work input are found equal in that machine
Explanation:
Monoammonium phosphate effectively smothers the fire, while sodium bicarbonate induces a chemical reaction which extinguishes the fire. Fire extinguishers with a Class C rating are suitable for fires in “live” electrical equipment.
Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)
