Answer:
D
Explanation:
they can see this along with many other fish.
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.
There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.
You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.
Mass m = 68 kg
center of gravity from his palms x = 0.7 m
center of gravity from his feet x ' = 1 m
forces exerted by the floor on his palms and feet are F and F ' respectively.
with respect to palms :---------------------
( F*0 ) - (W * x ) + [ F ' * (x+x') ] = 0
-mg*0.7 + F ' * 1.7 = 0 where W = weight = mg
F ' * 1.7 = mg * 0.7
F ' = mg * 0.7 / 1.7
= 68 *9.8 * ( 0.7 / 1.7 )
= 274.4 N
with respect to feet :--------------------
( F ' * 0 ) -( W* x ' ) + [F * ( x + x') ] = 0
-mg*1 + [ F * 1.7 ]= 0
F = mg / 1.7
= 392 N
Answer:
Explanation:
During a course of elliptical motion the direction of the body keeps changing however its magnitude may or may not remain constant.
During a curve path motion the velocity of the object is directed in a tangential direction to the curve.
During the motion on a curved path if the magnitude of a body remains constant then the body is said to have a constant speed which is a scalar quantity. Scalar quantities do not have direction.
