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3 years ago

Write the equality and conversion factors for the relationship between miles and hours for a car traveling at 55 mi/h

Physics

1 answer:

AysviL [449]3 years ago

5 0

Using Raoult’s Law, we can calculate for the vapor pressure using the formula:

P = x * P(pure)

Where x is the mole fraction of water, P(pure) = pressure of pure water, P = pressure when in solution

calculating for mole fraction of water:

x = (0.42kg / 18kg/kmol) / ((0.42kg / 18kg/kmol) + (0.075kg / 192.12kg/kmol))

x = 0.9835

Therefore:

P = 0.9835* P(71.93 mmHg)

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Consider four different types of electromagnetic radiation: microwaves, infrared, ultraviolet, and X–rays. Arrange the types of

oksian1 [2.3K]

Answer:

A) from the lowest to the highest frequency:

Microwave - infrared - ultraviolet- X-ray

B) from the highest to the lowest energy:

X-ray - ultraviolet - infrared - microwave

Explanation: electromagnetic spectrum is the range of all types of electromagnetic radiation.

Radiation is the energy that travels and spreads out as it goes out.

4 0

3 years ago

An electromagnetic wave is transporting energy in the negative y direction. At one point and one instant the magnetic field is i

natali 33 [55]

Answer:. Option c

Explanation: the speed of an electromagnetic wave is simply the vector product of the magnetic field and the electric field.

The direction of the velocity is the direction of the electromagnetic wave.

The wave is already moving towards the negative y axis (-j) and the magnetic field is already pointing towards the positive x axis (i)

From cross product of unit vectors

i × j = k

i × k = - j

With the second identity, we can see that the electric field will be pointing towards the positive of the x axis (k).

Option c is validated

4 0

3 years ago

A portable music player, operating with four 1.5 V cells connected in series, provides a resistance of 15 000 Ω. What amount of

nata0808 [166]

Answer:

Explanation:

Given that,

A portable music player is operating with 4 cell batteries connected in series, and each cell has a P.D of 1.5V.

Then,

Total potential difference is

P.D_total = V1 + V2 + V3 + V4

P.D_total = 1.5 + 1.5 + 1.5 + 1.4

P.D_total = 6V.

The music player provides a resistance of 15,000Ω

R = 15,000Ω

We want to find the current (I) flowing through the music player?

Using ohms law

V = IR

Where

V is the potential difference

I is the current

R is the resistance

Therefore,

I = V/R

I = 6 / 15,000

I = 4 × 10^-4 A

I = 0.4 × 10^-3 A

I = 0.4 mA.

So, 0.4mA is passing through the music player

7 0

3 years ago

A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air ha

Tamiku [17]

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

θ = sin⁻¹ n₂ / n₁

let's calculate

θ = sin⁻¹ (1.00 / 1.50)

θ = 41.8º

4 0

3 years ago

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago