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Dahasolnce [82]
2 years ago
7

Why is charcoal black?

Physics
1 answer:
WARRIOR [948]2 years ago
8 0
The answer is C. 

I just took the test!

Hope this helps. :)
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Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c
kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

\lambda = \frac{c}{f}

Substituting,

\lambda = \frac{344}{1250}

\lambda = 0.2752m

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

sin\theta = \frac{m \lambda}{W}

Where,

W is the width

m is the integer

\lambda the wavelenght

Re-arrange the expression,

\theta = sin^{-1} \frac{m\lambda}{W}

For m=1,

\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°

For m=2,

\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°

For m=3,

\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

5 0
2 years ago
determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the
Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0
3 years ago
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0
2 years ago
A 55-kg woman is wearing high heels. If each heel has a circular cross-section 6.0 mm in diameter and she puts all her weight on
bogdanovich [222]

Answer:

The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

Explanation:

Given;

mass of the woman, m = 55 kg

diameter of the circular heel, d = 6.0 mm

radius of the heel, r = 3.0 mm = 0.003 m

Cross-sectional area of the heel is given by;

A = πr²

A = π(0.003)²

A = 2.8278 x 10⁻⁵ m²

The weight of the woman is given by;

W = mg

W = 55 x 9.8

W = 539 N

The pressure exerted by the woman on the floor is given by;

P = F / A

P = W / A

P = 539 / (2.8278 x 10⁻⁵ )

P = 1.9061 x 10⁷ N/m²

Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

5 0
3 years ago
What is step one of a star called?​
yKpoI14uk [10]

Answer: Stage 1- Stars are born in a region of high density Nebula, and condenses into a huge globule of gas and dust and contracts under its own gravity. This image shows the Orion Nebula or M42 . Stage 2 - A region of condensing matter will begin to heat up and start to glow forming Protostars.

Explanation:

3 0
3 years ago
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