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.
K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2
So, option B is your answer!
Hope this helps!
Explanation:
The given data is as follows.
Length (l) = 2.4 m
Frequency (f) = 567 Hz
Formula to calculate the speed of a transverse wave is as follows.
f = 
Putting the gicven values into the above formula as follows.
f = 
567 Hz = 
v = 544.32 m/s
Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.
Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
The answer is A. Newton's third law of motion states that for every action, there is an equal and opposite reaction. A rocket exerts a large force on the gas that is in the rocket chamber (action). The gas thus exerts a large reaction force forward on the rocket (reaction). The large reaction force is called thrust.