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Likurg_2 [28]
3 years ago
10

The difference between experimental technique and procedure

Physics
1 answer:
kaheart [24]3 years ago
4 0
A procedure is all the steps used to do an experiment in order. 
<span>the experiment is when you test your hypothesis and is designed to answer your question. </span>
<span>the procedure is all the steps of the experiment.</span>
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The binding energies of K-shell and L-shell electrons in a certain metal are EK and EL, respectively, If a Kαx ray from this met
Svetach [21]

Answer:

The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

Explanation:

The relationship between energy and wavelength is expressed below:

E = hc/λ

λ = hc/EK - EL

Considering the condition of Bragg's law:

2dsinθ = mλ

For the first order Bragg's law of reflection:

2dsinθ = (1)λ

2dsinθ = hc/EK - EL

d = hc/2sinθ(EK - EL)

Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.

Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

5 0
3 years ago
Basketball player Darrell Griffith is on record as
Gre4nikov [31]

Explanation:

1.

We use the equation

h = \frac{gt^2}{2}, where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

2.

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

5 0
3 years ago
A rectangular plate, whose streamwise dimension (or chord c) is 0.2 m and whose width (or span b) is 1.8 m, is mounted in a wind
34kurt

Answer:

Answer  for the question is given in the attachment.

Explanation:

Download pdf
4 0
3 years ago
Instantaneous speed is measured
VMariaS [17]

Answer:

C. At a particular instant

Explanation:

Speed is the defined as the ratio between the distance covered by an object and the time taken:

v=\frac{d}{t}

where d is the distance and t the time.

However, there are two possible measurements of speed:

- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is

v=\frac{100 m}{5 s}=20 m/s

- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.

7 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
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