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leonid [27]
3 years ago
9

Please Help, I need to turn this in today

Physics
2 answers:
Scrat [10]3 years ago
4 0
You're right, its C. Good work
mart [117]3 years ago
3 0
I think it is C , idk why, but I just learn about this and I think it is.... sorry I didn't really help.....
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Please help solving this physics question
sp2606 [1]
True
False
False.
True
True
False
4 0
3 years ago
A spherical gas-storage tank with an inside diameter of 9 m is being constructed to store gas under an internal pressure of 1.50
lutik1710 [3]

Answer: 33 mm

Explanation:

Given

Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m

Internal pressure of gas, P(i) = 1.5 MPa

Yield strength of steel, P(y) = 340 MPa

Factor of safety = 0.3

Allowable stress = 340 * 0.3 = 102 MPa

σ = pr / 2t, where

σ = allowable stress

p = internal pressure

r = radius of the tank

t = minimum wall thickness

t = pr / 2σ

t = 1.5*10^6 * 4.5 / 2 * 102*10^6

t = 0.033 m

t = 33 mm

The minimum thickness of the wall required is therefore, 33 mm

6 0
3 years ago
Please help ASAP! Thank you :)
puteri [66]

Answer:

magnitude of gravitational force between the Earth and the Sun at B is greater than that at A

Explanation:

Formula of gravitational force:

F = GMm/r^2

(r is the distance between 2 objects)

We see that r(B) < r(A) since at B, the Earth is closer to the Sun than at A

According to the Formula, the smaller r is, the greater F is

So, F(B) > F(A)

8 0
3 years ago
A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change
notsponge [240]
We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
where
\Delta U is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ
6 0
3 years ago
What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
PolarNik [594]

Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m

So,

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

Also, \Delta E=h\times \nu

So,

h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}

\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

8 0
3 years ago
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