The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
The question can be solved, using Newton's second law of motion.
Note: Momentum of the cannon = momentum of the cannonball.
<h3>
Formula:</h3>
- MV = mv................. Equation 1
<h3>Where:</h3>
- M = mass of the cannon
- m = mass of the cannonball
- V = velocity of the cannon
- v = velocity of the cannonball
Make v the subject of the equation.
- v = MV/m................ Equation 2
From the question,
<h3>Given: </h3>
- M = 500 kg
- V = 3 m/s
- m = 10 kg.
Substitute these values into equation 2.
- v = (500×3)/10
- v = 150 m/s.
Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
Learn more about Newton's second law here: brainly.com/question/25545050
Answer:
Explanation:
Speed given = 125 m /min
125 /60 m /s
In 450 second it will travel
= 450 x 125 / 60
=937.5 m.
As the distance is covered in less than 450 seconds , The distance must be less than 937.5 m
In 400 seconds , it will travel
= 400 x 125 / 60
833.33 m
Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .
Hence the distance covered is more than .833 m but less than 937.5
In either case these distance are more than .8 km .
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Answer:
(C) The frequency decrease and intensity decrease
Explanation:
The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.
if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.
The frequency is the inverse from the wavelength, so the frequency heard will increase.
The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.
Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513
