Identify two types of nuclear reactions in which the equations E= mc^2 applies.
1 answer:
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Answer:
d) None of the above
Explanation:
= inituial velocity of launch = 4 m/s
θ = angle of launch = 10 deg
Consider the motion along the vertical direction
= initial velocity along vertical direction = 4 Sin10 = 0.695
m/s
= acceleration along the vertical direction = - 9.8 m/s²
y = vertical displacement = - 20 m
t = time of travel
using the equation
- 20 = (0.695) t + (0.5) (- 9.8) t²
t = 2.1 sec
consider the motion along the horizontal direction
x = horizontal displacement
= initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s
= acceleration along the horizontal direction = 0 m/s²
t = time of travel = 2.1 s
Using the kinematics equation
x = (3.94) (2.1) + (0.5) (0) (2.1)²
x = 8.3 m
Consider the motion along the vertical direction
= initial velocity along vertical direction = 4 Sin10 = 0.695
m/s
= acceleration along the vertical direction = - 9.8 m/s²
=initial vertical position at the time of launch = 20 m
= vertical position at the maximum height = 20 m
= final velocity along vertical direction at highest point = 0 m/s
using the equation
= 20.02 m
h = height above the starting height
h = -
h = 20.02 - 20
h = 0.02 m
Answer:
Vertical velocity decreases.
Explanation:
The motion of the ball is a projectile ball, which consists of two independent motions:
- a horizontal motion, with constant velocity
- a vertical motion, with constant acceleration g=9.8 m/s^2 towards the ground
In the vertical motion, there is a constant acceleration directed downward: this means that the vertical velocity decreases as the ball goes higher. In fact, it decreases following the equation
And it decreases until the ball reaches its maximum height, then it starts increasing again.