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wlad13 [49]
2 years ago
12

What is represent by number 14​

Physics
2 answers:
Maru [420]2 years ago
7 0
None are correct ,it shows the atomic number
otez555 [7]2 years ago
6 0

Answer:

C po baka namn nakatulong

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Please help with vectors (will give BRAINLIEST answer)
zhenek [66]

just analyze it in this way:

20cos30*=10( radical 3 )

20sin30*=10

7 0
3 years ago
If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi
saul85 [17]
The gravitational potential energy:
E p = m x g x h
where m is the mass and h is the height;
m = 50 kg,  g = 9.81 m/s² ,  h = 40 m
E p = 50 kg  x 9.81 m/s²  x  40 m
Answer:
E p = 19,620 J = 19.62 kJ
8 0
2 years ago
Read 2 more answers
COULD SOMEONE PLS DO THESE FOR ME <br> I’LL LOVE U FOREVER AND EVER
sladkih [1.3K]
I can help you do them for yourself :)
7 0
3 years ago
Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
BigorU [14]

Answer:

Explanation:

Since energy is conserved:

2

mu  

2

 

​

=  

2

mv  

2

 

​

+mgh

⇒u  

2

=v  

2

+2gh

⇒(3)  

2

=v  

2

+2(9.8)(0.5−0.5cos60)

⇒v=2m/s

7 0
3 years ago
Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black
oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

Where,

F =View Factor

A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75

Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

\dot{Q_3} = 780.76W

Therefore the rate heat radiation is 780.76W

5 0
2 years ago
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