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3 years ago

Photochemical smog consists of

Physics

1 answer:

musickatia [10]3 years ago

3 0

It is a mixture of sunlight and nitrous oxides that leaves airborne particles in the atmosphere

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What is used as evidence for sea-floor spreading?

raketka [301]

Answer:

Sea-floor spreading occurs in the oceanic ridges. In there, volcanic activity, together with the gradual movement of the bottom, form new oceanic crust. This allows a better understanding of the continental drift explained by the theory of plate tectonics.

The greatest evidence for Sea-floor spreading is the oceanic trenches, the oceanic ridges, the magma protruding to the surface and the new seafloor.

In previous theories, continents were assumed to be transported across the sea. Harry Hess, in the 1960s, proposed the idea that the seabed itself moves as it expands from a central point. The theory is now accepted, and the phenomenon is thought to be caused by convection currents in the upper layer of the mantle.

4 0

4 years ago

Read 2 more answers

A tennis ball is dropped from 1.43 m above the

Rudiy27

Answer:

-5.29 m/s

Explanation:

Given:

y₀ = 1.43 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.43 m)

v = -5.29 m/s

4 0

4 years ago

A sprinter accelerates at 7.5 m/s from rest in 2.0 s, what distance did she go? (15 m)

a_sh-v [17]

Answer:

<em>The sprinter traveled a distance of 7.5 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the rate of change of the velocity of an object is constant.

The equation that rules the change of velocities is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:

$\displaystyle a=\frac{7.5-0}{2}$

$a=3.75\ m/s^2$

Now calculate the distance:

$\displaystyle x=0*2+\frac{3.75*2^2}{2}$

$\displaystyle x=7.5\ m$

The sprinter traveled a distance of 7.5 m

8 0

3 years ago

Select the correct answer.

Gelneren [198K]

Answer: It's b and c I got it right

Explanation:

Hope this helped!!!! :)

8 0

3 years ago

a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi

zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the given values in the above equation

a = (10 - 0) / 3

= 3.33 m/s²

Using the second equations of motion

s = ut + 1/2 at²

= 0 + 0.5 x 3.33 x 3²

= 15 m

The force exerted by the car

F = m x a

= 1500 Kg x 3.33 m/s²

= 5000 N

The work done by the car,

W = F x S

= 5000 N x 15 m

= 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

P = W / t

= 75,000 J / 3 s

= 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0

4 years ago