Answer:
(a) Angular acceleration is 1.112 rad/s².
(b) Average angular velocity is 2.78 rad/s .
Explanation:
The equation of motion in Rotational kinematics is:
θ = θ₀ + 0.5αt²
Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.
(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:
13.9 = 0 + 0.5α(5)²
α = 1.112 rad/s²
(b) The equation of average angular velocity is:
ω = Δθ/Δt
ω = 
ω = 2.78 rad/s
Answer:
tan 249 = 2.61
tan 249 = tan (249 - 180) = tan 69 = 2.61
Answer:
<u><em>Rate of dissolving compounds:</em></u>
If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.
<u><em>Boiling Point of Compounds:</em></u>
If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
