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soldier1979 [14.2K]
2 years ago
10

Kate is researching air pollution and finds some information on ozone. She knows that ozone is a good thing as part of the ozone

layer, so why is it also an air pollutant?
A) At ground level, ozone contributes to smog.


B) We breathe out ozone, adding it to the atmosphere.


C) Ozone breaks into carbon dioxide.


D) It contributes to global warming.
Physics
1 answer:
jek_recluse [69]2 years ago
7 0

"At ground level, ozone contributes to smog" so it is also an air pollutant.

Option: A

<u>Explanation</u>:

ozone is naturally present in stratosphere and acts as shield against harmful ultraviolet radiations. But it acts a pollutant contributing to global warming when it is present in lower level atmosphere particularly troposphere. In this level it combines with primary pollutants that is "nitrogen oxides" and "volatile organic" compounds to form secondary pollutant which absorbs outgoing radiation and contributes in raising the temperature. It has harmful impacts on vegetation as well as human health.

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Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,
Alex Ar [27]
Ans: 
1.  τbiceps = +(Positive)
2.  τforearm = -(Negative)
3.  τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0
2 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
1 year ago
Pions have a half-life of 1.8 x 10^-8 s. A pion beam leaves an accelerator at a speed of 0.8c. What is the expected distance ove
Nuetrik [128]

Answer:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × 10^{-8} s

speed = 0.8 c = 0.8 × 3 × 10^{8}

to find out

expected distance over

solution

we know c is speed of light in air is 3 × 10^{8} m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × 10^{-8} ×  0.8 × 3 × 10^{8}

expected distance = 4.32

so the expected distance is 4.32 m

5 0
2 years ago
What will be the result of sea level rising, causing the ocean to fill a glacially carved valley?
Alex
Global warming is what will happen
7 0
2 years ago
The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2
masha68 [24]

Answer:

the normal stress induced by  the concrete post \sigma_c = 67.26 psi

the normal stress induced by the steel \sigma_s = - 1795.84 psi

Explanation:

Given that:

Modulus for elasticity for concrete post E_c = 3.6 *10^6 psi

Thermal coefficient for concrete post  \alpha _c = 5.5 *10^{-6}/^0F

Modulus for elasticity of steel bar E_s = 29*10^6psi

Thermal coefficient of steel bar \alpha _2 = 6.5*10^{-6}/^0F

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod A_s = 6(\frac{ \pi}{4} )(d_s)^2

= 6(\frac{ \pi}{4} )(\frac{7}{8} )^2

= 3.61 in²

Area of concrete parts A_c = (10)(10) - A_s

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= [\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)

= [\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)

= [(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}

= 1.234*10^{-8}P = 8.0*10^{-5}

P = \frac {8.0*10^{-5}}{1.234*10^{-8}}

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

\sigma_c =\frac{P}{A_c}

\sigma_c = \frac{6482.98}{96.39}

\sigma_c = 67.26 psi

Thus, the normal stress induced by  the concrete post \sigma_c = 67.26 psi

Also; the normal stress in the steel bars  induced as a result of temperature rise is as follows:

\sigma_s = \frac{-P}{A_s}

\sigma_s =\frac{-6482.98}{3.61}

\sigma_s = - 1795.84 psi

Thus, the normal stress induced by the steel \sigma_s = - 1795.84 psi

6 0
2 years ago
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