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Semenov [28]
2 years ago
6

Sound travels at a speed of about 344 m/s in air. You see a distant flash of lighting and hear the thunder arrive 7.6 seconds la

ter. How many miles away was the lighting strike? (assume the light takes essentially no time to reach you.)
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Mazyrski [523]2 years ago
8 0

Answer:

0.056 miles away

Explanation:

From sound wave,

v = 2x/t .................................. Equation 1

Where v = velocity of sound in air, x = distance of echo, t = time.

making x the subject of the equation,

x = 2v/t........................... Equation 2.

Given: v = 344 m/s, t = 7.6 s.

Substituting into equation 2

x = 2(344)/7.6

x = 90.53 m.

x = 90.53/1609.344

x = 0.056 mile.

Thus the lighting strike 0.056 miles away

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A slanted vector has a magnitude of 41 N and is at an angle of 23 degrees north of east. What are the magnitude and direction of
jenyasd209 [6]

Answer:

The horizontal component is 37.74 N to east

The vertical component is 16.02 N to North

Explanation:

Any slant vector has two components

→ Horizontal component = R cos Ф

→ Vertical component = R sin Ф

→ R is the magnitude of the vector

→ Ф is the direction of the vector with positive part of the horizontal axis

A slanted vector has a magnitude of 41 N and is at an angle of 23

degrees north of east

23° north of east means the angle between the vector and the

east direction is 23° (east is the positive horizontal direction)

That means R is 41 N and Ф is 23°

→ R = 41 N , Ф = 23°

→ The horizontal component = 41 × cos(23) = 37.74 N east

→ The vertical component = 41 × sin(23) = 16.02 N North

<em>The horizontal component is 37.74 N to east</em>

<em>The vertical component is 16.02 N to North</em>

7 0
3 years ago
A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

5 0
3 years ago
Read 2 more answers
IN THIS FORMULA FOR WATER WHAT DOES THE SUBSCRIPT 2 INDICATE
masha68 [24]
The chemical formular for water is H2O.
The H aspect of the formula stands for hydrogen gas and the subscript 2 which is attached to the H symbol signifies that two atoms of hydrogen are joined together, that is two atom of hydrogen are present.
The chemical formula of water indicates that, two atom of hydrogen react with one atom of oxygen to form one molecule of water.
In chemical formulae, subscripts are normally used to indicate the number of atoms that are present in a molecule.
7 0
3 years ago
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
2 years ago
What does strong language mean ?
Art [367]

Answer:

noun

the style of a piece of writing or speech.

"he explained the procedure in simple, everyday language"

coarse or offensive language.

noun: strong language

"strong language"

Explanation:

comment how it helps

3 0
3 years ago
Read 2 more answers
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