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3 years ago

A gas can compress and change shape easily. How do these properties relate to the motion of atoms in a gas?

Physics

1 answer:

Ksju [112]3 years ago

3 0

Answer:

The atoms in a gas move about freely and have a lot of space between them.

Explanation:

Just took the test on edge 2020

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An ice cube at 0°C is placed in a very large bathtub filled with water at 30°C and allowed to melt, causing no appreciable chang

Paladinen [302]

Answer:

C. The entropy of the system (ice plus water) increases because the process is irreversible.

6 0

3 years ago

Explain when acceleration remains constant.

stellarik [79]

Explanation:

acceleration remains constant when velocity does not increase or decrease

6 0

3 years ago

Build a second circuit with a battery and a light bulb but this time add a switch. Your circuit might look something like the on

kompoz [17]

Answer:

the resistance of the wire has no effect on the brightness of the bulb.

Explanation:

Let's apply ohm's law for your light bulb circuit plus wires plus switch

V = I R_{bulb} + I R_ {wire}

the current in a series circuit is constant

V = I (R_{bulb} + R_{wire})

To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.

Incandescent bulb

Power 60 W

let's use the power ratio

P = V I = V2 / R

R = V2 / P

the voltage value for this power is V = 120 V

R = 120 2/60

R_bulb = 240 Ω

Resistance of a 14 gauge copper wire (most used), we look for it on the internet

R = 8.45 Ω/ km

in a laboratory circuit approximately 2 m is used, so the resistance of our cable is

R = 8.45 10⁻³ 2

R_wire = 0.0169 Ω

let's buy the two resistors

R_{bulb} = 240

R_{wire} = 0.0169

$\frac{R_{bulb} }{R_{wire} } = \frac{240 }{ 0.0169}$

$\frac{ R_{bulb} }{ R_{wire} } = 1.4 \ 10^4$

therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.

In summary, the resistance of the wire has no effect on the brightness of the bulb.

6 0

3 years ago

A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced sli

goblinko [34]

Given Information:

Wavelength of the red laser = λr = 632.8 nm

Distance between bright fringes due to red laser = yr = 5 mm

Distance between bright fringes due to laser pointer = yp = 5.14 mm

Required Information:

Wavelength of the laser pointer = λp = ?

Answer:

Wavelength of the laser pointer = λp = ?

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

y = Dλ/d

y/λ = D/d

Where

λ is the wavelength

y is the distance between bright fringes.

d is the double slit separation distance

D is the distance from the slits to the screen

For the red laser,

yr/λr = D/d

For the laser pointer,

yp/λp = D/d

Equating both equations yields,

yr/λr = yp/λp

Re-arrange for λp

λp = yp*λr/yr

λp = (5*632.8)/5.14

λp = 615.56 nm

Therefore, the wavelength of the small laser pointer is 615.56 nm.

3 0

3 years ago

Wonder Woman and Superman fly to an altitude of 1690 km , carrying between them a chest full of jewels that they intend to put i

Arturiano [62]

Answer:

5120 m/s

Explanation:

The acceleration due to gravity is:

g = MG / r²

where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.

Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.

g = MG / (h + R)²

Acceleration is the derivative of velocity:

dv/dt = MG / (h + R)²

Using chain rule, we can say:

(dv/dh) (dh/dt) = MG / (h + R)²

(dv/dh) v = MG / (h + R)²

Separate the variables:

v dv = MG / (h + R)² dh

Integrating:

∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²

½ v² |₀ᵛ = -MG / (h + R) |₀ʰ

½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)

½ v² = -MG / (h + R) + MG / R

½ v² = MGh / (R(h + R))

v² = 2MGh / (R(h + R))

Given:

M = 5.98×10²⁴ kg

R = 6.37×10⁶ m

h = 1.69×10⁶ m

G = 6.67×10⁻¹¹ m³/kg/s²

Plugging in:

v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))

v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷

v ≈ 5120 m/s

Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)

v ≈ 5760 m/s

So we know that our calculated velocity of 5120 m/s is a reasonable answer.

3 0

3 years ago