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alexandr1967 [171]
2 years ago
12

Most houses in the countries that experience winter have air heaters installed.These heaters are usually installed near the floo

r.Explain why.
Physics
1 answer:
Tatiana [17]2 years ago
6 0

Answer: because the air currents move upwards so the hot air fills up the whole room. Similarly if the air conditioner is placed neer the ceiling because it gives cool air

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What is the SI unit of pressure? A. g/cm3 B. the pascal C. the newton D. m/s2
sesenic [268]
The basic definition of pressure is force/area and the scientific community defined that as the Pascal (Pa).
6 0
3 years ago
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Consider two diffraction gratings with the same slit separation. The only difference between the two gratings is that one gratin
kobusy [5.1K]

Answer:

True The grid with more slits gives more angle separation increases

True. The grating with 10 slits produces better-defined (narrower) peaks

Explanation:

Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is

      d sin θ = m λ

where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.

For network with 5 slits

      d = 1/5 = 0.2

For the network with 10 slits

      d = 1/10 = 0.1

let's calculate the separation (teat) for each one

      θ = sin⁻¹ (m λ / d)

for 5 slits

     θ₅ = sin⁻¹ (m λ 5)

for 10 slits

     θ₁₀ = sin⁻¹ (m λ 10)

we can appreciate that for more slits the angle increases

the intensity of a series of slits is

       I = I₀ sin²2 (N d/2) / sin² d/2)

when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)

let's analyze the claims

False

True The grid with more slits gives more angle separation increases

False

True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases

False

5 0
3 years ago
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might
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<span>Density is 3.4x10^18 kg/m^3 Dime weighs 1.5x10^12 pounds The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so 4/3 pi 1.9x10^3 = 4/3 pi 6.859x10^3 m^3 = 2.873x10^10 m^3 Now divide the mass by the volume 9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3 Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3 Now to figure out how much the dime weighs, just multiply by the volume of the dime. 3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg And to convert from kg to lbs, multiply by 2.20462, so 6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
4 0
3 years ago
This famous scientist conducted experiments that led to the development of the Plum Pudding atomic model.
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J.J. Thomson discovered the electron, leading to the Plum Pudding model
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3 years ago
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