Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:
Where,
x is in meters and t is in sec
We know that,
Velocity,
(a) i. t = 2 s
At t = 4 s
(b) Acceleration,
Pu t = 3 s in the above equation
So,
Hence, this is the required solution.
Answer:
1.5 times
Explanation:
= depth of the diver initially = 5 m
= density of seawater = 1030 kg m⁻³
= Initial pressure at the depth
= final pressure after rising = 101325 Pa
Initial pressure at the depth is given as
= Initial volume at the depth
= Final volume after rising
Since the temperature remains constant, we have
Answer:
-8.04 m/s2
Explanation:
To find the answer to this, you have to use the 4th kinematic equation:
You plug into the equation to get:
solve for a to get
-8.04 m/s2
460 meters per second, or about 1,000miles per hour.
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
