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pickupchik [31]
2 years ago
10

Two beams of coherent light are shining on the same piece of white paper. with respect to the crests and troughs of such waves,

darkness will occur on the paper where
Physics
1 answer:
defon2 years ago
7 0

Answer:

"where crests and troughs have their maxima at the same time"

Crests and troughs are 180 deg out of phase and when they have their maxima at the same time and place, their net contribution will be zero"

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Imagine you have a collection of identical flat-bottomed coffee filters that can be nested (stacked inside of each other) so tha
skad [1K]

Answer:

the terminal velocity of 14 nested coffee filters is 3.2 m/s

Explanation:

Given the data in the question;

we know that;

The terminal velocity is proportional to the square root of weight.

v ∝ √W

v = k√W

the proportionality constant depends upon the surface area and the density of the medium (like air). The coffee filters can be stacked such that the resulting area is roughly unchanged. So, the constant of proportionality k is also unchanged

v/√W = constant

v₂/√W₂ = v₁/√W₁

v₂ = v₁√(W₂ / W₁ )

given that;

v₁ = 0.856 m/s,

W₂ = 14W₁; meaning 14 coffee filters have 14 times the weight of a single coffee filter

so we substitute

v₂ = 0.856 √(14W₁  / W₁ )

v₂  = 0.856 √( 14( W₁/W₁)

v₂  = 0.856 √( 14(1)

v₂  = 0.856 √( 14 )

v₂  = 0.856 × 3.741657

v₂  = 3.2 m/s

Therefore, the terminal velocity of 14 nested coffee filters is 3.2 m/s

6 0
2 years ago
Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What
yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

<u>Step 1: </u>Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water  =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

<u>Step 2:</u> formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

<u>Step 3: </u>Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

<u>Step 4: </u>Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

<u>Step 5:</u> Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

<u>Step 6</u>: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

<u>Step 7:</u> Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0
2 years ago
Is this charging by induction or conduction?
sineoko [7]
Conduction i believe
3 0
2 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
Clois what is the weight of a body in the earth, if its weig is 5Nin moon? <br>​
Debora [2.8K]

Explanation:

because the moon has less mass than earth, the force due to gravity at the lunar surface is only about 1/6 that on earthso,the weight of a body on earth is 6×5N =30N

8 0
2 years ago
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