(a) The lunch angle is 64.8⁰
(b) The initial speed of the pass when the angle of projection is 25⁰ is 21.2 m/s
(c) The time of flight of the bullet is 1.83 s
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The given parameters include;
time of flight, T = 3.97 s
initial velocity, u = 21.5 m/s
(a) The lunch angle is calculated from the equation of motion of time of flight;
(b) the initial speed of the pass when the angle of projection is 25⁰ and range of 35 m, is calculated as follows;
(c) The time of flight of the bullet is calculated as;
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Answer:
Mass of water,
Explanation:
Given that,
Surface area of the reservoir, A = 30 km² = 3 × 10⁷ m²
Average depth, d = 49 m
The volume of the reservoir is V such that,
We have to find the mass of water is held behind the dam. It can be calculated using the expression for density. We know that density of water, d = 1000 kg/m³
Density,
Hence, this is the required solution.
Answer: D
Explanation: Look at the figure, we can conclude that the correct answer is
X: Low potential energy
Y: High potential energy
Z: Flow of electrons
Because electrons flow where there is difference in potential energy. And electrons move from a region of high potential energy to a region of low potential energy.
Since the arrow is pointing to X, that means
Z is the flow of electrons, X is of low potential energy and Y is of high potential energy.
Answer:
(a) 0.277 m
(b) 0.553 m
Explanation:
(a) For the closed cylinder,
Applying,
f' = v/4l............. Equation 1
Where f' = fundamental frequency, v = velocity of sound, l = lenght of the cylinder
make l the subject of the equation
l = v/4f'............. Equation 2
From the question,
Given: v = 343 m/s, f' = 310 Hz
Substitute these values into equation 2
l = 343/(4×310)
l = 0.277 m
(b) For the open cylinder,
f' = v/2l
l = v/2f'................ Equation 3
Substitute into equation 3
l = 343/(2×310)
l = 0.553 m