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
Answer:
Explanation:
The question relates to time of flight of a projectile .
Time of flight = 2 u sinθ / g
u is speed of projectile , θ is angle of projectile
= 2 x 48.5 sin42 / 9.8
= 6.6 seconds .
Maximum height attained
= u² sin²θ / g
= 48.5² sin²42 / 9.8
= 107.47 m .
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
A) 1 rev = 2π rad. Using this ratio, you can find the rad/s: 1160 rev/min x 2π rad/rev x 1 min/60 s = 121.5 rad/s
b) You can find linear speed from angular speed using this equation (note the radius is half the diameter given in the question): v = ωr = 121.5 rad/s x 1.175 m = 142.8 m/s
c) You can find centripetal acceleration using this equation: a = v^2/r = (142.8 m/s)^2 / 1.175 m = 17 355 m/s^2
