Supposing there's no air
resistance, horizontal velocity is constant, which makes it very easy to solve
for the amount of time that the rock was in the air.
Initial horizontal
velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s
15.5m / 10.3923m/s = 1.49s
So the rock was in the air for 1.49 seconds. </span>
<span>
Now that we know that, we can use the following kinematics
equation:
d = v i * t + 1/2 * a * t^2
Where d is the difference in y position, t is the time that
the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>
<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s
So:
d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89</span>
d = -1.95<span>
<span>This means that the initial y position is 1.95 m higher than
where the rock lands. </span></span>
Answer:
the power that can be generated by the river is 117.6 MW
Explanation:
Given that;
Volume flow rate of river v = 240 m³/s
Height above the lake surface a h = 50 m
Amount of power can be generated from this river water after the dam is filled = ?
Now the collected water in the dam contains potential energy which is used for the power generation,
hence, total mechanical energy is due to potential energy alone.
= m(gh)
first we determine the mass flow rate of the fluid m
m = p×v
where p is density ( 1000 kg/m³
so we substitute
m = 1000kg/m³ × 240 m³/s
m = 240000 kg/s
so we plug in our values into (
= m(gh) kJ/kg )
= 240000 × 9.8 × 50
= 117600000 W
= 117.6 MW
Therefore, the power that can be generated by the river is 117.6 MW
Electric potential energy of a dipole is given as


so change in the potential energy is given by


here initially it was parallel to electric field while finally it is perpendicular to the electric field
so we have


so above is the change in potential energy of dipole
Answer:
jsjdgsudgwid s
Explanation:
83638hr is 3738 so 8273 and 837y37 and 82638 say hi to 1937
Answer:
D. circumpolar
Explanation:
A circumpolar star is a star, as viewed from a given latitude on Earth, that never sets below the horizon due to its apparent proximity to one of the celestial poles.
-Wikipedia