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.
Answer:
The maximum permissible propagation delay per flip flop stage is<u> 100 </u>n sec
Explanation:
1024 ripple counter has 10 J-K flip flops(210 = 1024).
So the total delay will be 10×x where x is the delay of each J-K flip flops.
The period of the clock pulse is 1× 10⁻⁶ s.
Now
10x <= 10⁻⁶ s
x <= 100 ns
x= 100 ns for prpoer operation.
pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operation of the counter, the maximum permissible propagation delay per flip flop stage is <u>100 </u>n sec.
Momentum definition: the quantity of motion of a moving body, measured as a product of its mass and velocity.
Impulse definition: change of momentum of an object when the object is acted upon by a force for an interval of time
These are the definitions.
Hope this helps!
- Melanie
Answer:
1) Yes, the work done by gravity during the trip upward is -41.405·m J
2) Yes, the work done by gravity on the ball during the trip downward = 41.405·m J
Explanation:
The given parameters are;
The initial velocity of the ball on the way up = 9.1 m/s
The final velocity of the ball on the way up = 0 m/s
The initial velocity of the ball on its trip downward = 0 m/s
The final velocity of the ball on its trip downward = 9.1 m/s
1) The work done by gravity on the trip upward = The change in kinetic energy of the ball
The change in kinetic energy of the ball = 1/2 × m × 9.1² = -41.405·m J, given that the initial kinetic energy is 41.405·m J and the final kinetic energy is J
Where;
m = The mass of the ball
Therefore, the work done by gravity during the trip upward = -41.405·m J
2) Similarly gravity does work on the ball during the trip downward = 41.405·m J.
