I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
I think god did ??? I searched it up okay
The zone that gases always accelerate upward is the Luminous flame zone. The fire plume is the column of hot gases, flames and smoke rising above a fire. Gases accelerate upward toward the always luminous flame zone. The luminous flame height is the distance between the base of a flame and the point at which the plume is luminous half the time and transparent half the time.
Answer:
Option (D) is correct.
Explanation:
The balloon lands horizontally at a distance of 420 m from a point where it as released.
Velocity of air balloon along +X axis =10 m/s
velocity of ball=4 m/s along + X axis
the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s
time taken= 30 s
The distance traveled is given by d= v t
d= 14 (30)
d= 420 m
Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.
