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miv72 [106K]
2 years ago
15

HELP ASAPP

Physics
1 answer:
Travka [436]2 years ago
3 0
Radioactive dating :)
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One scientist suggests that out of the different possible locations, they should design the model and build it at the equator re
Anika [276]

Answer:

B.

I think.

Explanation:

Mars doesn't have that much of an atmosphere!

Have a great day!

7 0
3 years ago
Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH = 1.06
Slav-nsk [51]

Answer:

1.06085\times 10^{-10}\ m

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

m = Mass of electron = 9.11\times 10^{-31}\ kg

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

e = Charge of electron = 1.6\times 10^{-19}\ C

n = 1 (ground state)

Angular momentum is given by

L=mvr

From Bohr's atomic model we have

L=\dfrac{nh}{2\pi}

mvr=\dfrac{nh}{2\pi}\\\Rightarrow v=\dfrac{nh}{2\pi mr}

The centripetal force will balance the electrostatic force

\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow \dfrac{ke^2}{r}=mv^2\\\Rightarrow \dfrac{ke^2}{r}=m(\dfrac{nh}{2\pi mr})^2\\\Rightarrow r=\dfrac{n^2h^2}{4\pi^2mke^2}\\\Rightarrow r=\dfrac{1^2\times (6.626\times 10^{-34})^2}{4\pi^2 \times 9.11\times 10^{-31}\times 8.99\times 10^{9}\times (1.6\times 10^{-19})^2}\\\Rightarrow r=5.30426\times 10^{-11}\ m

The diameter is 2\times 5.30426\times 10^{-11}=1.06085\times 10^{-10}\ m

7 0
3 years ago
The velocity of an object is the _____ of the object
katovenus [111]
D. Speed and direction, this is because velocity is a vector quantity so has a magnitude and direction assigned to it because it is the rate of change of displacement.
6 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line
Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

y=\frac{m\lambda D}{d}

where

m is the order of the maximum

\lambda is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

\lambda=632.8 nm = 632.8\cdot 10^{-9} m is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

n=520 lines/mm is the number of lines per mm, so the spacing between two lines is

d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m

Therefore, substituting m = 1, we find:

y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m

So, on the distant screen, there is 1 bright fringe every 33 cm.

6 0
3 years ago
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