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Mnenie
2 months ago
11

In aviation, it is helpful for pilots to know the cloud ceiling, which is the distance between the ground and lowest cloud. The

simplest way to measure this is by using a spotlight to shine a beam of light up at the clouds and measuring the angle between the ground and where the beam hits the clouds. If the spotlight on the ground is 0.75 km from the hangar door. What is the cloud ceiling?
Physics
1 answer:
Sav [3.1K]2 months ago
4 0

Answer:

b = a \frac{cos(\alpha)}{sin(\alpha)} = a Cot(\alpha)

Explanation:

To find the height of the beam where it strikes the clouds, apply the Pythagorean theorem using these formulas:

b  = \sqrt[2]{H^{2} - a^{2}}

b = \sqrt[2]{H^{2} - H^{2} sin^{2} (\alpha)} \\b = H \sqrt[2]{1 - sin^{2} (\alpha)}

If:

H = \frac{a}{sin(\alpha)}

Then:

b = \frac{a}{sin(\alpha)} \sqrt{1 - sin^{2}(\alpha)}\\b = \frac{a}{sin(\alpha)} \sqrt{1 - 1 + cos^{2}(\alpha)}\\

b = \frac{a}{sin(\alpha)} \sqrt{cos^{2}(\alpha)}\\b = a \frac{cos(\alpha)}{sin(\alpha)} = a Cot(\alpha)

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Imagine you are a water molecule, carbon atom, or nitrogen atom that is inside a cow. Write a brief explanation of how you got f
Maru [3345]

Answer:

Explanation:

Each of the processes connected to these molecules varies.

For instance, water that has accumulated in the atmosphere returns to the ground as rain. Cows utilize this water from local water sources. This represents one method in which water transitions from the atmosphere to the cow's body.

Regarding carbon and nitrogen, the air inhaled by cows contains nitrogen, oxygen, carbon dioxide, and other gases. These molecules enter the cow through respiration.

6 0
1 month ago
1. Use Coulomb’s Law (equation below) to calculate the approximate force felt by an electron at point A in the schematic below.
Keith_Richards [3271]

Answer:

Explanation:

The data indicates that point A is located midway between two charges.

To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

This field points towards Q⁻.

A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

For the force acting on an electron placed at A:

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

8 0
1 month ago
Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w
Keith_Richards [3271]

Solution:

Em_{f} / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

      K = ½ m v²

      U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

    Em₀ = K = ½ m v²

At its highest point

    Em_{f} = U = mg and

Now let's compute

    Em₀ = ½ m 3.6²

    Em₀ = m 6.48

    Em_{f} = m 9.8 × 0.2

    Em_{f} = m 1.96

Thus the energy lost is given by:

    Em_{f} / Em₀ = m 1.96 / m 6.48

   Em_{f} / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

8 0
1 month ago
Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
Ostrovityanka [3204]

Answer:

2.32\times 10^{-11}

Explanation:

The first number is 4.48\times 10^{-8}.

The second number is 5.2\times 10^{-4}.

We must multiply these two numbers together.

4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}

In scientific notation: 2.32\times 10^{-11}

Therefore, this is the solution you are looking for.

8 0
2 months ago
Two frictionless lab carts start from rest and are pushed along a level surface by a constant force. Students measure the magnit
Yuliya22 [3333]

Answer:

The kinetic energy is higher for the first cart.

Explanation:

For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to

(2kg)v = 10kg\: m/s

resulting in

v = 5m/s.

Consequently, the kinetic energy for the 3kg cart ends up as

K.E.  = \dfrac{1}{2}mv^2

= \dfrac{1}{2}(2kg)(5m/s)^2 =  25J

\boxed{K.E = 25J}

indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.

3 0
2 months ago
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