PHYSICS HIGH SCHOOL

If you divide the mass of a substance by its volume, you can find its density. Please select the best answer from the choices provideda. True
b. False

Answers

Answer 1
Answer:

Answer: True

Explanation: density is a measure of the mass of matter contained by a unit volume. So, dividing the mass in grams, of any substance by its volume in centimeter cube, gives it's density in grams per centimeter cube (g/cm³)

Answer 2
Answer: It is True. To find the density you must divide the mass by the volume 

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Alternative between running snd walking

Answers

this would have to be jogging

Mary walked north from her home to Sheila's home, which is 4.0 kilometers away. Then she turned right and walked another 3.0 kilometers to the supermarket, which is 5.0 kilometers from her own home. She walked the total distance in 1.5 hours. What were her average speed and average velocity?

Answers

It would be 5.666667 km per hour.
well her speed and velocity are the same 8 kilometers per hour

1. Coherent beams with wavelengths of 400 nm intersect at points on the screen. What will be observed at these points interference maximum or minimum, if the optical path difference of the beam is in the 1st case A1=20 um and in the 2nd - A2=15 um? Find the order of minimum and maximum. 2. An installation for observing Newton's rings in reflected light is illuminated by monochromatic light incident normally to the plate surface. The distance between the second and fourth dark rings is 5 mm. Find the distance between the eighth and sixteenth light rings. 3. Find the angle between the main planes of the polarizer and the analyzer if the intensity of the natural light passing through the polarizer and the analyzer decreased by 5 times.

Answers

  • 1. The observed interference at points will be maximum for \(A_1 = 20 \, \mu m\) and minimum for \(A_2 = 15 \, \mu m\), with orders 12th maximum and 12th minimum.
  • 2. The distance between the eighth and sixteenth light rings is approximately 2.583 mm.
  • 3. The angle between the main planes of the polarizer and the analyzer is approximately \(63.43^\circ\) when the intensity decreases by 5 times.

1. For coherentbeams with wavelengths of 400 nm, the interference observed at points on the screen will be maximum when the optical path difference (OPD) is a whole number of wavelengths (\(m \lambda\)), and minimum when the OPD is a half-integer number of wavelengths (\((m + 0.5) \lambda\)).

Given

\(A_1 = 20 \, \mu m\)

\(A_2 = 15 \, \mu m\), the difference in optical path between the two cases is \(\Delta A = A_1 - A_2 = 5 \, \mu m\).

The number of wavelengths difference can be calculated as \((\Delta A)/(\lambda)\), which is approximately 12.5 wavelengths.

2. In Newton's ringspattern, the radius of the m dark ring is given by \(r_m = √(m \lambda R)\), where R is the radius of curvature of the lens.

Given the distance between the second and fourth darkrings (\(m = 2\) to \(m = 4\)) as 5 mm, we can set up an equation:

\(√(4 \lambda R) - √(2 \lambda R) = 5 \, \text{mm}\).

Solving for R gives \(R \approx 1.904 * 10^(-2) \, \text{m}\).

Now, the radius of the 16th light ring can be calculated as \(r_(16) = √(16 \lambda R)\), and the radius of the 8th light ring is \(r_(8) = √(8 \lambda R)\). The distance between these two light rings is then given by:

\(d = r_(16) - r_(8) \n\n= √(16 \lambda R) - √(8 \lambda R)\).

Plugging in the values, we get \(d \approx 2.583 \, \text{mm}\).

3. When the intensity decreases by a factor of 5, the transmission of the polarizer and analyzer combination becomes \((1)/(√(5))\) times the original transmission, as intensity is proportional to the square of the transmission.

The relationship between the intensity and the angle \(\theta\) between the main planes of the polarizer and the analyzer is given by Malus's law: \(I = I_0 \cos^2 \theta\), where \(I_0\) is the initial intensity.

Given that \(I_0\) decreases by a factor of 5, we have:

\((I)/(I_0) = (1)/(√(5)) \n\n= \cos^2 \theta\)

Solving for \(\theta\), we get:

\(\theta = \arccos \left( (1)/(√(5)) \right)\n\n \approx 63.43^\circ\)

Thus, the angle is \(63.43^\circ\).

For more details regarding wavelength, visit:

brainly.com/question/31143857

#SPJ4

Answer:

a=3....................................

Ix identical square pyramids can fill the same volume as a cube with the same base. If the height of the cube is h units, what is true about the height of each pyramid? a.The height of each pyramid is h units.
b.The height of each pyramid is h units.
c.The height of each pyramid is h units.
d.The height of each pyramid is h units.

Answers

Answer:

height of each pyramid must be 3h units

Explanation:

As we know that the volume of the cube is same as that of volume of square pyramid

so we have volume of the cube is given as

V = h^3

now if the base area of pyramid and the cube is same

so we will have

V_(pyramid) = (1)/(3) Base\: Area * height

V_(pyramid) = (1)/(3) h^2 * y

now since both have same volume

(1)/(3)h^2 y = h^3

y = 3h

height of each pyramid must be 3h units

Ball X is dropped from a height of 1.25 meters above the ground. At the same time and from exactly the same height, Ball Y is shot outhorizontally at an initial velocity of 8.0 m/s. Air resistance is negligible and assume g =-10 m/s?
How far away from Ball X will Ball Y hit the ground?

Answers

Answer: The distance between Ball X and Ball Y is 4.0 meters

Explanation:

First we have to use the information we have from Ball Y to find time as well as it’s distance.

Ball Y’s Given:

Vx=8.0m/s

ax=0m/s^2

x=?

y=1.25m

ay=-10m/s^2

To find the time, use the equation t=(2y/g)^1/2.

t=(2(1.25)/10)^1/2

t=(2.5/10)^1/2

t=(0.25)^1/2

t=0.5s

Now we have to calculate how far away Ball Y fell from the base of the cliff. To do this use the equation x=(Vx)(t) and plug in the values.

x=(8.0m/s)(0.5s)

x=4.0m

Because Ball X is dropped from the cliff, we can conclude that the ball is 0meters away from the cliff. With this in mind, we can assume that the distance between Ball X and Ball Y is 4.0m.

Total Distance = Ball Y - Ball X

Total Distance = 4.0 m - 0 m

Total Distance = 4.0 m

In a series circuit, as more batteries are added what effect does it have on the bulbs in a circuit . Will the bulbs get brighter or stay the same?

Answers

If the batteries are in series too, then the more batteries,
the more voltage across the light bulbs, and the brighter
they will all burn. 

(Until you add the battery that causes the weakest bulb
to burn out.  Then they ALL go out.)
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