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3 years ago

Particle accelerators make it easy for scientists to study what?

2 answers:

6 0

A particle accelerator is a machine that uses electromagnetic fields to propel charged particles to nearly light speed and to contain them in well-defined beams. The answer is C

monitta3 years ago

3 0

The answer is electricity

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Quick does the watermelon have more or less mass then the 2kg bottle?

Ratling [72]

Answer

it will be less\

Explanation:

6 0

3 years ago

Select all of the statements that are true.

Strike441 [17]

I think its all four of them could be wrong but try all four !!!!!!

7 0

3 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

kogti [31]

Answer:

a) $y_{first}=5.3mm$

b) $y_{second}=10.6-5.3 =5.3 mm$

Explanation:

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

m is 1 (to find the central bright fringe)
D is the distance from the slit to the screen
a is the slit wide
λ is the wavelength

So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

$y_{first}=5.3mm$

Now, if we do m=2 we can find the distance to the second minima.

$y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}$

$y_{2}=10.6 mm$

Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

4 0

3 years ago

After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter

Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, $\Delta L=3.1\ cm$

(a) The change in length per unit original length is called strain. So,

$\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193$

(b) The force acting per unit area is called stress.

$\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa$

$Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2$

Hence, this is the required solution.

8 0

3 years ago

The force between two point-charges of 0.040 C and 0.060 C separated by certain distance is 2.2 x 10^5 N. what is the distance b

Gala2k [10]

Answer:

im in hardvard dont cheat and get good grades and become a graduate

it is c

Explanation:

8 0

3 years ago

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