<h2>Answer:</h2>
The distance will be 1800 m
<h2>Explanation</h2>As in question
Time = 15 min
Time = 15 x 60 sec = 900 sec
Speed = 2 m/s
We know that
So, the answer is 1800 m
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
1 answer:
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
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Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
(-2, -1)(3, 1)
Therefore,
1) 
<Solve using the formula which is:
Mass=Density×Volume
2)
3)
(Length × Width × Height)
(Answer=7.5)
Mass=Density×Volume
2)
3)