This is also called a plum-pudding model of an atom which has positive charges spread throughout a spherical figure with electrons embedded uniformly within its background.
Hope this answers your question! :)
Answer:
If the amplitude of motion is doubled, so the mass swings 8 cm to one side and then the other, the period of the motion will be <u>2 s</u>.
Explanation:
As we know that time period of the motion of the simple pendulum is given as
here we know that
L = length of the pendulum
g = acceleration due to gravity
so as per above formula we know that time period of the pendulum is independent of the amplitude of the motion of the time period
So we will say that there is no change in the time period of the motion when we increase the amplitude of the motion of the given pendulum
Answer:
Id = 1/2 Md * R^2 = 1/2 * .1 * .1^2 = .0005 kg m^2 inertia of disk
Ib = Mb * R^2 = .02 * .1^2 = .0002 kg m^2 inertia of bug at edge
(Id + Ib) w1 = Id w2 conservation of angular momentum
w2 = .0007 / .0005 * 10 = 14 /sec angular speed with bug at center
KE1 = 1/2 I1 * w1^2 = 1/2 * .0007 * 10^2 = .035 kg m^2 / s^2
KE2 = 1/2 * I2 w2*2 = (.0005 / 2 ) ^ 14^2 = .049 kg m^2 / s^2
The bug has to exert radial force on the disk to maintain its
centripetal acceleration. As the bug crawls to the center of the disk it
does work against this centripetal force which appears as an increase
of rotational energy of the disk. As the the bug crawls back to the edge
of the disk, the disk does work on the bug and loses KE.
Answer:
a 10.6 m/s²
Explanation:
Since F = ma (Force = mass * acceleration), acceleration would be...
a = F/m
a = 302 N/28.6
a 10.6 m/s²
Answer:
The gauge pressure is 1511.11 psi.
Explanation:
Given that,
Flow rate = 94 ft³/min
Diameter d₁=3.3 inch
Diameter d₂ = 5.2 inch
Pressure P₁= 15 psi
We need to calculate the pressure on other side
Using Bernoulli equation
We know that,
Where, V = volume
v = velocity
A = area
Put the value of v into the formula
Put the value into the formula
We need to calculate the gauge pressure
Using formula of gauge pressure
Put the value into the formula
Hence, The gauge pressure is 1511.11 psi.