False, sharks and remoras have a symbiotic relationship. The remora removes parasites from the sharks scales, and the shark provides protection for the remora
Answer:
D...............................
Convex mirrors only produce virtual images
<span>3.78 m
Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes.
7.2 m/s / 9.81 m/s^2 = 0.77945 s
The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving
d = 1/2 A T^2
d = 1/2 9.81 m/s^2 (0.77945 s)^2
d = 4.905 m/s^2 0.607542 s^2
d = 2.979995 m
So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height.
d = 2.979995 m + 0.8 m = 3.779995 m
Rounding to 2 decimal places gives us 3.78 m</span>
Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N