1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kotykmax [81]
3 years ago
12

What type of smog is created when

Physics
1 answer:
Airida [17]3 years ago
5 0

Answer:

d seems like the best answer to but I don't know.

Explanation:

good luck and sorry if its incorrect

You might be interested in
Which would be the best example to demonstrate the principle of conservation of energy? A) a stone is crushed B) light bends aro
Daniel [21]
The answer is a pendulum swings back an forth.
6 0
3 years ago
Read 2 more answers
In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.
HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is 41.9^{\circ}

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

s=u_y t + \frac{1}{2}at^2 (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

u_y = u sin \theta is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

\theta=40.0^{\circ} is the angle of projection

So

u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s

a=g=-9.8 m/s^2 is the acceleration due to gravity (downward)

Substituting the numbers, we get

-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

d=u_x t

where

u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

d=(9.19)(1.778)=16.34 m

B)

In this second case,

\theta=42.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s

So the equation for the vertical motion becomes

4.9t^2-8.1t-1.80=0

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s

So, the range of the shot is

d=u_x t = (8.85)(1.851)=16.38 m

C)

In this third case,

\theta=45^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s

So the equation for the vertical motion becomes

4.9t^2-8.5t-1.80=0

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s

So, the range of the shot is

d=u_x t = (8.49)(1.925)=16.34 m

D)

In this 4th case,

\theta=47.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s

So the equation for the vertical motion becomes

4.9t^2-8.8t-1.80=0

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s

So, the range of the shot is

d=u_x t = (8.11)(1.981)=16.07 m

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is \theta=42.5^{\circ}.

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

s-u sin \theta t + \frac{1}{2}gt^2=0

The solutions of this quadratic equation are

t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}

This is the time of flight: so, the horizontal range is

d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta

It can be found that the maximum of this function is obtained when the angle is

\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})

Therefore in this problem, the angle which leads to the maximum range is

\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
If the wavelength of a wave increases does its frequency also increase?
irinina [24]
Frequency decreases whilst wavelength increases and the opposite also occurs
4 0
3 years ago
What is the maximum range of most handheld fire extinguishers?        A. 50 yd   B. 30 ft   C. 10 ft   D. 100 ft
alex41 [277]
C- 10ft. Hope this helped. Have a great day! :D
5 0
3 years ago
Read 2 more answers
What is accurate about the planet’s climate system?
ElenaW [278]

The accurate about the planet’s climate system is the wind

because heating near the equator blows the wind to drive the convection cells in the atmosphere, and the friction created by the rotation of the spherical planet in the atmosphere causes the wind to appear to bend left or right across the surface of the planet. ..

The climate system is a highly complex global system consisting of five major components: the atmosphere, the ocean, the cryosphere (cryosphere), the land surface, the biosphere, and the interactions between them.

Solar energy drives the climate by heating the surface of the earth unevenly. Ice also reflects incoming sunlight, further cooling the poles. Temperature differences move the ocean and atmosphere as they work together to disperse heat throughout the globe.

Learn more about the planet’s climate system here:brainly.com/question/15351986

#SPJ4

6 0
2 years ago
Other questions:
  • Find the force needed to accelerate a .3 kg bullet at 2100 m / s / s.
    8·1 answer
  • How much does a 0.15 kg baseball weigh on earth?
    11·1 answer
  • Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you.
    13·2 answers
  • Suppose you increase your walking speed from 5 m/s to 14 m/s in a period of 3 s. What is your acceleration
    12·1 answer
  • A person pushes a box across the floor the energy from the person moving arm is transferred to the box in the box in the floor b
    12·1 answer
  • A force of 20 N acted on a 5 kg object, moving it a distance of 4 m. How much work was done on the object?
    7·1 answer
  • Question 1 of 6
    6·1 answer
  • A 70 kg bicyclist rides his 9.8 kg bicycle with a speed
    13·1 answer
  • Help me <img src="https://tex.z-dn.net/?f=%5Chuge%5Cmathcal%5Cred%7B%7D" id="TexFormula1" title="\huge\mathcal\red{}" alt="\huge
    13·1 answer
  • 7
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!