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3 years ago

13

There are slightly more than one hundred elements known to exist. The smallest complete unit of an element that serves as the ba

sic building block of matter is the:
A) atom
B) Proton
C) neutron
D) nucleus

1 answer:

Nostrana [21]3 years ago

7 0

A) it is atom for e2020

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The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?

belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

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2 years ago

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A thin flake of mica (n=1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the wiew

Llana [10]

To develop this problem it is necessary to apply the optical concepts related to the phase difference between two or more materials.

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

Where

L = Thickness

n = Index of refraction of each material

$\lambda =$ Wavelength

Our values are given as

$\Phi = 7(2\pi)$

$L=t$

$n_1 = 1.58$

$n_2 = 1$

$\lambda = 550nm$

Replacing our values at the previous equation we have

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

$7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))$

$t = \frac{7*550}{1.58-1}$

$t = 6637.931nm \approx 6.64\mu m$

Therefore the thickness of the mica is 6.64μm

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2 years ago

Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

vlabodo [156]

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2 years ago

How to measure the volume of a baseball bat ( need answers ASAP )

vaieri [72.5K]

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water. But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly. Some water will run over the top and out of the tub. That's OK ... it's exactly what you want. If NO water runs over the top, pour some more in, until it runs out and then stops. You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan. The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string. Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it. The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub. Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub. Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it. Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions. Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces. But you can find all those conversion factors with a search on Floogle.

8 0

2 years ago

A mass of 1 kg is moving in a circle of radius 3.3 meters. What is the liner velocity v m/s that would give a Centripetal Force

Ghella [55]

Answer:

The linear velocity of the object is 8.71 m/s.

Explanation:

Given;

mass of the object, m = 1 kg

radius of the circle, r = 3.3 meters

centripetal force, F = 23 N

Centripetal force is given by;

$F_c = \frac{mv^2}{r}\\\\$

where;

v is the linear velocity of the object

$F_c = \frac{mv^2}{r}\\\\mv^2 = F_cr\\\\v^2 = \frac{F_cr}{m}\\\\v= \sqrt{\frac{F_cr}{m}} \\\\v= \sqrt{\frac{23*3.3}{1}}\\\\v = 8.71 \ m/s$

Therefore, the linear velocity of the object is 8.71 m/s.

4 0

3 years ago