Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=
⇒ v=
\
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N
Answer:
An example of a weak base is ammonia. It does not contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. The position of equilibrium varies from base to base when a weak base reacts with water. The further to the left it is, the weaker the base.
Explanation:
I hope this helps!!
Answer:
PE is related to the ability to do work, If an item is sitting on a shelf
it has potential energy relative to its position on the floor, However, if the object were to fall it would hit the floor with a KE equal to the PE that it had sitting on the shelf.
Sounds are caused by compressional waves in the air - when a piano key is struck or a TV is turned on, then compressional waves are produced in the surrounding air due to a disturbance. The human ear recognizes the disturbed air as due to the object that created the disturbance.
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa