The height of the projectile is given by:
h = ut - 0.5at²
The height will be 0 twice, once at t = 0 and the second time at the time when the journey has been completed, or t = hang time.
0 = 7.3t - 0.5(9.81)t²
t = 1.49 seconds
Momentum = (mass) x (velocity)
Original momentum before the hit =
(0.16 kg) x (38 m/s) this way <==
= 6.08 kg-m/s this way <==
Momentum after the hit =
(0.16) x (44 m/s) that way ==>
= 7.04 kg-m/s that way ==>
Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .
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Change in momentum = impulse.
Impulse = (force) x (time the force lasted)
13.12 kg-m/s = (force) x (0.002 sec)
(13.12 kg-m/s) / (0.002 sec) = Force
6,560 kg-m/s² = 6,560 Newtons = Force
( about 1,475 pounds ! ! ! )
"The speed will increase" is the one among the following choices given in the question that describes the speed of a wave traveling through the rope, if <span>the tension in a rope is increased. The correct option among all the options that are given in the question is the first option. I hope that this is the answer that has come to your help.</span>
Answer:
The grating spacing of the beetle is 
Explanation:
The concept to solve this problem is relate to interference effect given in the Young's Slits. Here was demonstrated that the length of the side labelled \lambda is known as the path difference. The equation is given by,
Where,
= wavelenght of light
N = a positive integer: 1,2,3...
= Angle from the center of the wall to the dark spot
d= width of the slit
Replacing our values we have that for n=1,
Therefore the grating spacing of the beetle is
Answer;
=15855.40 kg/m^3
Explanation;
Volume (V) of the cylinder = pi x r^2 x h
V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3
V = 6.307 x 10^-5 m^3
By density = m/V
mass = 1 kg
density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3
