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svp [43]
3 years ago
9

Hey Tori's are the rim of a canyon yells hello toward the opposite side she has an echo hello four seconds later if the speed of

sound is 340 m/s how far away was the canyon wall that produced the echo?
Physics
1 answer:
krek1111 [17]3 years ago
6 0

Answer:

The wall is 680 meter away from the person.

Explanation:

Given data

Speed of sound = 340 \frac{m}{s}

Given that Persons said hello toward the opposite side she has an echo hello 4 seconds later means it takes 2 seconds for the sound to reach the wall & again 2 seconds to reach the persons ear.

Therefore the distance between the person & wall is

D = speed × Time

D = 340 × 2

D = 680 meter

Therefore the wall is 680 meter away from the person.

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PART A)

By Snell's law we know that

n_1sin i = n_2 sin r

here we know that

n_1 = 1.5

i = 25 degree

n_2 = 1

now from above equation we have

1.5 sin25 = 1 sin r

r = 39.3 degree

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see  that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}

d_i = 30 cm

so image will form at 30 cm from mirror

it is virtual image and smaller in size

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Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!
DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

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During science class, while studying mixtures, you mix together iron filings and sand. Your teacher challenges you to separate t
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We can say that we can use the magnet to attract the iron filings out of the mixture because iron is magnetic solid, but sand will not attract because sand is not magnetic solid.

So,  we use a magnet to pull out the iron filings as they are attracted to a magnet.

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