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
Answer:
(a). The frequency of this standing wave is 0.782 kHz.
(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.
Explanation:
Given that,
Length of tube = 11.0 cm
(a). We need to calculate the frequency of this standing wave
Using formula of fundamental frequency

Put the value into the formula



(b). If the test tube is half filled with water
When the tube is half filled the effective length of the tube is halved
We need to calculate the frequency
Using formula of fundamental frequency of the fundamental standing wave in the air

Put the value into the formula



Hence, (a). The frequency of this standing wave is 0.782 kHz.
(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.
Given that,
Mass of the object, m = 1 kg
It moves in a circle of radius 0.5 m with a speed of 5 m/s
To find,
The direction of the acceleration.
Solution,
Whenever an object moves in a circular path, the only force that acts on its is centripetal force which is given by the formula as follows :

The centripetal acceleration acts in the direction of force. It acts along the radius of the circular path. In this figure, the direction of the acceleration is shown at point d.