Answer:
Explanation:
As we know the , equation of time period for simple pendulum ,
T = 2*pi*
hence putting values we get ,
the solution is in picture ,
please
Brain-list it or support me at my U-Tube channel " ZK SOFT&GAMING " I will be thankful
We do not feel the gravitational forces from objects other than the Earth because they are weak.
Answer:
d) -4.0
Explanation:
The magnification of a lens is given by
where
M is the magnification
q is the distance of the image from the lens
p is the distance of the object from the lens
In this problem, we have
p = 50 cm is the distance of the object from the lens
q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct
Also, q is positive since the image is real
So, the magnification is
Answer: you want your input force harder
Explanation:
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
