sound travels through air at around 340 - 350 m/s.
to cover 52 meters takes it, say, (52/345) = about 0.15 second
This would be the same time-lag even if it were not a major-league game,
except that you might not be able to get a seat 52 meters from home plate
at a little league or t-ball game.
Answer:
Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.
Explanation:
Answer:
Not be changed
Option: D
<u>Explanation:</u>
The physical quantity which has both ‘magnitude and direction’ is called vector. These vectors are represented by a line and an arrow, <em>the line represent the magnitude and arrow represent the direction of the physical quantity</em>. The vectors are added and subtracted according to the direction of the vectors.
According to the vector law addition while adding vectors direction and length of the vector is not be changed.<em> If the length of the vector changed the magnitude is also changed while so, while adding vectors length must not be changed.
</em>
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.
