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ludmilkaskok [199]
2 years ago
12

Solar energy breaks oxygen molecules apart in the stratosphere, releasing ____, which combine(s) with additional oxygen molecule

s to form ____.
Physics
1 answer:
Pepsi [2]2 years ago
5 0

Answer:

Oxygen atom

Ozone Molecule

Explanation:

The upper stratosphere and Mesosphere consist of the ozone layer. This layer helps in protecting us from the UV radiations of the Sun. This layer converts the UV radiation into heat energy.

The UV radiation breaks down the Oxygen molecules in oxygen atom which further combines with oxygen molecule to form Ozone. This Ozone molecule breaks to form Oxygen atom and Oxygen molecule. Oxygen atom further recombines with Oxygen molecule to make Ozone molecule. This is known as Ozone-Oxygen cycle. The excess energy is released as heat.

Everyday around 400 million metric tons of Ozone is produced through this cycle.

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a hiker walks 200 m west and then walks 100 m north in what direction is her resulting displacement draw to show your answer​
Sliva [168]

Answer:

The direction of the displacement is in North-West.

Explanation:

Resultant displacement D is

=\sqrt{(200)^{2} + (100)^{2} } \\=223.60m                        

Here the direction is

\Theta =tan^{^{-1}}\left ( \frac{100}{200} \right )\\\Theta =26.6^{o}

Then the direction is 26.6^{o} North-west.            

5 0
2 years ago
A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.
aleksley [76]
Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
                            d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
                          d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters. 
3 0
2 years ago
A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori
yarga [219]
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.


First, the vertical component of tension (Tsin theta) is equal to the weight of the object. 
 T * sin θ = mg =</span> 1.55 * 9.81 <span>
 T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind. 
 T * cos θ =  13.3 

Tan θ = sin  </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82. 

T then is equal to 20.20 N
4 0
3 years ago
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
Which terrestrial ecosystem or life zone produces the highest net primary productivity per year? a. temperate forest b. Savanna
horsena [70]

a. hope this helps...

8 0
3 years ago
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