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3 years ago

Harvesting timber (wood) from a forest does not harm___________.

2 answers:

8 0

It does not harm the buildings or the cars in a nearby city. However, it does harm just about everything in the forest that's anywhere near the (former) tree.

Trava [24]3 years ago

6 0

Your answer would be soil. When you harvest timber (wood), you affect the wildlife around it, not the soil. Cutting down a tree won't make much of a difference to the soil underneath or around it.

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A brand of earplugs reduces the sound intensity level by 27 dB.By what factor do these earplugs reduce the acoustic intensity?

Fofino [41]

Answer:

Explanation:

27dB = 2.7 B

So I / I₀ = 10⁻²°⁷ , I₀ is intensity of main sound and I is intensity after reduction.

= 1.99 X 10⁻³

So intensity will reduce by 1.99 X 10⁻³ .

3 0

3 years ago

What are the two ways that scientists can study earth's history

Alika [10]

If you mean climate change. Then scientists can study it by seeing where places and things are eroded.

8 0

3 years ago

Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

#SPJ4

3 0

2 years ago

If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100

balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s = (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0

3 years ago

Pahelp po ako.

NeTakaya

D bc thats sound like the only resonable answer

4 0

2 years ago