Answer:
The kinetic energy will have an increase as it falls.
Explanation:
The energy of a particle falling freely will be the sum of its kinetic energy and its potential energy.
When the pig is just dropped, the total enrgy will be only its potential energy. As it falls down its stored potetial energy will be converted into kineitic energy. Just before it touches the trampoline its total energy will be solely kinteic energy.
Thus, the kinetic energy will have an increase as it falls.
Answer:
A light emitting diode (a semiconductor diode which glows when voltage is applied.
Answer:
a) 15.49
b) Opposite to the ball's initial velocity
c) 258.16N
Explanation:
a)
b)
Since the player is kicking the ball in the opposite direction to which it came, the impulse is being directed opposite to the ball's initial velocity.
c)
Hope this helps!
The centripetal force : F = 293.3125 N
<h3>Further explanation</h3>
Given
mass = 65 kg
v = 9.5 m/s
r = 20 m
Required
the centripetal force
Solution
Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle
F = centripetal force, N
m = mass, Kg
v = linear velocity, m / s
r = radius, m
Input the value :
F = 65 x 9.5² / 20
F = 293.3125 N
This question is incomplete, the complete question is;
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass
;
a) F_throw = 8.083 N
b) F_throw = 9.181 N
c) F_throw = 2.284 N
d) F_throw = 16.014 N
e) None of these is correct
Answer:
the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Explanation:
Given that;
m = 0.408 kg
d = 1.909 m
u = 0 { from rest}
t = 0.439 s
Now using Kinetic equation
d = ut + 1/2 at²
we substitute
1.909 = (0 × 0.439) + 1/2 a(0.439)²
1.909 = 0 + 0.09636a
1.909 = 0.09636a
a = 1.909 / 0.09636
a = 19.8111 m/s²
Now force applied will be;
F = ma
we substitute
F = 0.408 × 19.8111
F = 8.0828 ≈ 8.083 N
Therefore the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
