Answer:
877
Explanation:
P1 = 150 atm
V1 = 0.12 m^3
P2 = 1.2 atm
V2 = ?
radius of each balloon, r = 0.16 m
Volume of each balloon,
![V_{2}=\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=V_%7B2%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
![V_{2}=\frac{4}{3}\pi 0.16^{3}](https://tex.z-dn.net/?f=V_%7B2%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%200.16%5E%7B3%7D)
V2 = 0.0171 m^3
Let the number of balloons be N.
So,
P1 x V1 = N x P2 x V2
150 x 0.12 = N x 1.2 x 0.0171
N = 877.19
So, the number of balloons be 877.
Im not sure but i guess you should try A
Gases have three characteristic properties:
(1) they are easy to compress,
(2) they expand to fill<span> their containers, and
(3) they occupy far more space than the liquids or solids from which they form.</span>
4m/s due north of the store
Explanation:
Given parameters:
Time taken = 45s
Displacement = 180m
Unknown
Velocity = ?
Solution:
Velocity is the displacement of a body per unit time. Velocity is a vector quantity that has both magnitude and direction.
Velocity =
= ![\frac{180}{45}](https://tex.z-dn.net/?f=%5Cfrac%7B180%7D%7B45%7D)
Velocity = 4m/s due north of the store
Learn more:
Velocity brainly.com/question/10883914
#learnwithBrainly
Answer:
R = 0.237 m
Explanation:
To realize this problem we must calculate the moment of inertia of the wheel formed by a thin circular ring plus the two bars with an axis that passes through its center.
The moments of inertia of the bodies are additive quantities whereby we can add the mounts inertia of the ring and the two bars.
Moment of inertia ring I1 = MR²
Moment of inertia bar I2 = 1/12 ML²
Moment of inertia disk I3 = ½ mR²
Let's calculate the moment of inertia of the wheel
I = I1 + 2 I2
I = MR² + 2 1/12 ML²
The length of the bar is ring diameter
L = 2R
I = 5.65 0.156² + 1/6 9.95 (2 0.156)²
I = 0.1375 + 0.1614
I = 0.2989 kg m²
This is the same moment of inertia of the solid disk,
Disk
I3 = I
I3 = ½ MR²
They give us disk density
ρ = M / V
M = ρ V
M = ρ (pi R² e)
Done is the thickness of the disc, in general it is e= 1 cm = 0.01 m
Let's replace
I3 = ½ ( ρ π R²) R²
I3 = ½ ρ π e R⁴
R⁴ = 2 I3 / ( ρ π e)
R = ( 2 I3 / ( ρ π e)
![)^{1/4}](https://tex.z-dn.net/?f=%29%5E%7B1%2F4%7D)
R⁴ = 2 0.2989 / (5990 π 0.01)
R = 0.237 m