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trapecia [35]
3 years ago
9

Which description BEST explains the direction of the moving force of air?

Physics
1 answer:
marin [14]3 years ago
3 0
Out of the choices given, the best choice to explain the direction of the moving force of air is from area o high pressure to areas of low pressure. 

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A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.
Neko [114]
In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
P_{2} = \frac{ P_{1}  T_{2} }{ T_{1} } = \frac{40*305}{289}  =42.21 psi
3 0
3 years ago
Why is no image formed when an object is at the focal point of a converging lens?
Temka [501]

Answer:

the refracted rays neither converge nor diverge. After refracting, the light rays are traveling parallel to each other and cannot produce an image.

Explanation:

6 0
3 years ago
6th grade science !!!!!!!!!!!!!!!​
posledela

Answer:

71

Explanation:

5 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
Alla [95]

Answer:

v_{ox}= 19.6\ m/s

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = v_{ox}

Initial vertical velocity= v_{oy}  (Since ball was thrown horizontally only)

Acceleration acting horizontally, a_x = 0 m/s²  [ Since no acceleration acts horizontally) ]

Vertical Acceleration, a_y = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

H= v_{oy}t+\frac{1}{2}a_yt^2

5= (0)t+\frac{1}{2}(9.8)t^2

t= \frac{10}{9.8}=1.02 s

Then using Equations of motion for horizontal motion,

R= v_{ox}t+\frac{1}{2}a_xt^2

20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2

v_{ox}= 19.6\ m/s

4 0
3 years ago
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