Answer:
F = 1.06N
Explanation:
In order to calculate the force that produces the given pressure you take into account the following formula:
(1)
P: pressure = 6.0kPa = 6.0*10³ Pa
F: force = ?
A: cross-sectional area of the tube = π r^2
r: radius of the tube = 1.5cm/2 = 0.75cm = 0.75*10^-2 m
You solve the equation (1) for F, and replace the values for the area A and the pressure P:
The force tha produces a pressure of 6.0kPa in a tube with a diameters of 1.5cm is 1.06N
Answer:
ω=6684.51 rpm
Explanation:
r= 30cm= 0.3m
a= 15000gs (convert to m/s^{2}
1g = 9.8 m/s^{2}
a= 15000 *9.8 = 147000 m/s^{2}
a=\frac{v^{2} }{r}
147000 = \frac{v^{2} }{0.3}
147000*0.3 = v^{2}
44100 = v^{2}
√44100 = v
210m/s = v
v=210m/s (linear velocity)
we will convert this to angular velocity
ω=\frac{v}{r}
ω= 210/0.3
ω= 700 rads^{-1}
we will convert this to rev per minute
1rad per second = 9.5493 rev per minute
ω= 700*9.5493
ω=6684.51 rpm
B is the answer. because it is a physical change
Answer:
the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019
Explanation:
given information
h = 0.2 m
= 1.01 x 
Pa
= + ρgh, ρ = 1000 kg/
= 1.01 x Pa + (1000 x 9.8 x 0.2) = 1,0296 x Pa
= = Pa
thus,
![\frac{V_{2} }{V_{1}} = 1,0296 x [tex]10^{5}](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B2%7D%20%7D%7BV_%7B1%7D%7D%20%3D%201%2C0296%20x%20%5Btex%5D10%5E%7B5%7D)
/ = 1.019
