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3 years ago

Explain how ozone in the atmosphere affects visible light on earth

1 answer:

8 0

Ozone gas in the upper atmosphere absorbs most of the ultraviolet radiation of wavelengths shorter than about 0.25 µm. This is actually a positive thing for us and most other living things, because of the harmful nature of ultraviolet radiation below these wavelengths. The ozone layer is a natural layer of gas in the upper atmosphere that protects humans and other living things from harmful ultraviolet (UV) radiation from the sun.

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The best way to choose the baseball bat that's right for you is by choosing one that is not too long and not too __________.

bearhunter [10]

Answer:

C. Heavy

Explanation:

You need to be able to lift the bat.

5 0

3 years ago

A woman can row a boat at 4.0 mph is still water.

vovikov84 [41]

Answer:

1) $\theta=120^{\circ}$ from the positive x-axis.

2) $t=20\ min$

Explanation:

Given:

speed of rowing in still water, $v=4\ mph$

speed of water stream, $v_s=2\ mph$

we know that the direction of resultant of the two vectors is given by:

$tan\ \beta=\frac{v.sin\ \theta}{v_s+v.cos\ \theta}$

where:

$\beta=$ the angle of resultant vector from the positive x-axis.

$\theta =$ angle between the given vectors

When the rower wants to reach at the opposite end then:

$\beta =90^{\circ}$

so,

$tan\ 90^{\circ}=\frac{v.sin\ \theta}{v_s+v.cos\ \theta}$

$\Rightarrow v_s+v.cos\ \theta=0$

$2+4\times cos\ \theta=0$

$cos\ \theta=-\frac{1}{2}$

$\theta=120^{\circ}$ from the positive x-axis.

Now the resultant velocity of rowing in the stream:

$v_r=\sqrt{v^2+v_s^2+2\times v.v_s.cos\ \theta}$

$v_r=\sqrt{4^2+2^2+2\times 4\times 2\times cos\ 120}$

$v_r=12\ mph$

Therefore time taken to cross a 4 miles wide river:

$t=\frac{4}{12}$

$t=\frac{1}{3}\ hr$

$t=20\ min$

8 0

3 years ago

5. Calculate how long in seconds it will take for a light pulse to travel from earth to the moon. The distance from earth to moo

solniwko [45]

Answer:

0.13 seconds

Explanation:

Since 1 Km = 0.621 miles

3.84 x 105 km = 3.84 x 105 × 0.621 = 23846.4 miles

Speed = distance/time

time= distance/speed

Time= 23846.4/186,000

Time= 0.13 seconds

3 0

3 years ago

The flux through the coils of a solenoid changes from 2.57.10-5 Wb to 9.44.10-5 Wb in 0.0154 s. If 4.08 V of EMF is generated, h

Vinil7 [7]

Hello!

We can use Faraday's Law of Electromagnetic Induction to solve.

$\epsilon = -N \frac{d\Phi_B}{dt}$

ε = Induced emf (4.08 V)
N = Number of loops (?)

$\Phi_B$ = Magnetic Flux (Wb)
t = time (s)

**Note: The negative sign can be disregarded for this situation. The sign simply shows how the induced emf OPPOSES the current.

Now, we know that $\frac{d\Phi_B}{dt}$ is analogous to the change in magnetic flux over change in time, or $\frac{\Delta \Phi_B}{\Delta t}$ , so:
$\epsilon = N \frac{\Delta \Phi_B}{\Delta t}\\\\\epsilon = N \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}$

Rearrange the equation to solve for 'N'.

$N = \frac{\epsilon}{ \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}}$

Plug in the given values to solve.

$N = \frac{4.08}{ \frac{9.44*10^{-5} - 2.57*10^{-5}}{0.0154}} = 914.585 = \boxed{915 \text{ coils}}$

**Rounding up because we cannot have a part of a loop.

4 0

2 years ago

A skydiver of mass m jumps from a hot air balloon and falls a distance d before reaching a terminal velocity of magnitude v. Ass

ahrayia [7]

Answer:

a) W = ΔK.E + ΔP.E = (mv²/2) - mgd

b) Power supplied by the drag force after the skydiver has reached terminal velocity = mgv

Explanation:

Using the work energy theorem, the work done by the drag force while the skydiver moves through a distance d is equal to the change in kinetic energy and potential energy of the body.

The potential energy of the skydiver drops by a magnitude of (mgd) falling through a distance of d.

ΔP.E = - mgd (skydiver loses this amount of potential energy)

But the skydiver falls from rest to gain a velocity of v after falling a distance of d vertically.

ΔK.E = (final kinetic energy at distance d) - (initial kinetic energy at rest)

Final kinetic energy at d = (1/2)(m)(v²) (the body is now falling with terminal velocity (v) at this point d)

Initial kinetic energy = 0 (since the skydiver was initially at rest)

ΔK.E = (1/2)(m)(v²) - 0 = (mv²/2)

W = ΔK.E + ΔP.E = (mv²/2) - mgd

b) At terminal velocity, the net force on the skydiver = 0

hence,

drag force = weight of the skydiver = mg

Power = F × v = mg × v = mgv

Hope this Helps!!!

5 0

3 years ago