Answer:
E = 10⁵ J
Explanation:
given,
Power, P = 100 TW
= 100 x 10¹² W
time, t = 1 ns
= 1 x 10⁻⁹ s
The energy of a single pulse is:-
Energy = Power x time
E = P t
E = 100 x 10¹² x 1 x 10⁻⁹
E = 10⁵ J
The energy contained in a single pulse is equal to 10⁵ J
Answer:
The maximum height reached by the body is 313.6 m
The time to return to its point of projection is 8 s.
Explanation:
Given;
initial velocity of the body, u = 78.4 m/s
at maximum height (h) the final velocity of the body (v) = 0
The following equation is applied to determine the maximum height reached by the body;
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u²/2g
h = (78.4²) / (2 x 9.8)
h = 313.6 m
The time to return to its point of projection is calculated as follows;
at maximum height, the final velocity becomes the initial velocity = 0
h = v + ¹/₂gt²
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
Answer:
Two stars, each of mass M, form a binary system. ... used is the distance between the centers of the planets; here that distance is 2R. ... r appears in the denominator of Newton's law of gravitation, the force of ... The orbital speed of a satellite orbiting the earth in a circular orbit at the ... is undergoing uniform circular motion?
Explanation:
Two stars, each of mass M, form a binary system. ... used is the distance between the centers of the planets; here that distance is 2R. ... r appears in the denominator of Newton's law of gravitation, the force of ... The orbital speed of a satellite orbiting the earth in a circular orbit at the ... is undergoing uniform circular motion?
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
