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Veseljchak [2.6K]
2 years ago
10

Light from a lamp is shining on a surface. how can you increase the intensity of the light on the surface?

Physics
1 answer:
vlada-n [284]2 years ago
4 0
The intensity of a light in a surface follows the inverse square law formula which can be mathematically expressed as,
                         I = k/d²
where I is intensity, d is distance, and k is the proportionality constant. For us to increase the intensity, we should lower the distance from the source to the surface. 
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Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s
USPshnik [31]
No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.
7 0
2 years ago
Read 2 more answers
Charge: A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (
snow_lady [41]

Answer

given,

net charge = +2.00 μC

we know,

1 coulomb charge =  6.28 x 10¹⁸electrons

1 micro coulomb  charge =  6.28 x 10¹⁸ x 10⁻⁶ electron

                                         = 6.28 x 10¹² electrons

2.00 μC = 2 x 6.28 x 10¹² electrons

              = 1.256 x 10¹³ electrons

since net charge is positive.

The number of protons should be 1.256 x 10¹³ more than electrons.

hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.

6 0
2 years ago
For what value m of the clockwise couple will the horizontal component ax of the pin reaction at a be zero? if a couple of that
sergejj [24]
Thank you for posting your question here at brainly. Below is the answer:

sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M 
<span>- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) </span>

<span>sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M </span>
<span>- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) </span>

<span>Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) </span>
<span>= (2M + 424.91)/(5.745) </span>

<span>M = ((275*5.745) - 424.91)/2 </span>
<span>= 577.483 which rounds off to 577 </span>

<span>Is it maybe supposed to be Ay - Cy = 275</span>
8 0
2 years ago
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If th
DiKsa [7]
<h2><u>Answer:</u></h2>

The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:

1. (There would be variables that would be hard to control, leading to less reliable data.)

3. (Meter sticks may lack precision or may be read incorrectly.)

4. (Real glider data may vary since real collisions may involve loss of energy.)

5. (Human error in recording or plotting the data could be a factor.)


6 0
2 years ago
Read 2 more answers
The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this
konstantin123 [22]

Answer:

3.81972\times 10^{-7}\ A

Explanation:

B = Magnetic field = 3\times 10^{-8}\ G

d = Diameter of loop = 16 cm

r = Radius = \frac{d}{2}=\frac{16}{2}=8\ cm

i = Current

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

The magnetic field of a loop is given by

B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A

The current needed to produce such a field at the center of the loop is 3.81972\times 10^{-7}\ A

5 0
2 years ago
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