Answer:
Liquid has a defined volume but undefined shape
Explanation:
Explanation:
Solid has specific shape and volume
Liquid has specific volume but no specific shape
Gases have no specific volume and no specific shape
So correct answer is liquid.
Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
Answer:
5.0 m/s
Explanation:
The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

where u is the initial speed and
. The horizontal distance travelled by the salmon is

where d = 1.95 m and t is the time needed to reach the final point.
Re-arranging for t,
(1)
Along the vertical direction, the equation of motion is

where:
y = 0.311 m is the final height reached by the salmon
h = 0 is the initial height
is the vertical component of the initial velocity of the salmon
is the acceleration of gravity
t is the time
Substituting t as found in eq.(1), we get the equation

and we can solve this formula for u, the initial speed of the salmon:

Kepler’s three law is the answer. Kepler’s 3 is the amount
of time it takes to orbit the sun is related to size and distance. Kepler’s 3 is one of the planetary motion and
can be stated as all planets move in elliptical orbits, having the sun sits at
one of the foci.
O.99 m long .simple pendulum time period is 2s for second formula then use formula T=2pi.rt(lenght/gravity)