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2 years ago

delivery ladies is Shivam noted at the thundering sound of 6 second after the lightning was seen by him

Physics

2 answers:

vitfil [10]2 years ago

8 0

Answer:

this is due to difference in speeds of sound and light

Explanation:

light has a speed of 3×10^8 m/s and it is seen at once because it takes negligible time due to very high speed and short distance that is why as soon as lightning occurs we can see it . since thundering sound travels with speed of sound which is about 330 to 340 m/s in air hence it takes some time as described in question as 6 seconds

Andre45 [30]2 years ago

7 0

Answer:

Explanation:

This is beacause light travel faster than sound

Lightening is light

Thunder is sound

That's why they first lightening and after6sec sound

Thank you

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An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act

jok3333 [9.3K]

Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector $\hat{i}$ .
The positive y-axis = the northern direction, with unit vector $\hat{j}$ .

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
$\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}$

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
$\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})$

The plane's actual velocity is the vector sum of the two velocities. It is
$\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}$

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.

5 0

2 years ago

You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class,

aksik [14]

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

$P_{r}=\dfrac{I}{c}$

$P_{r}=\dfrac{P}{Ac}Put the value into the formula[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}$

$P_{r}=2.75\ N/m^2$

We need to calculate the force

Using formula of force

$F=P\times A$

$F=P\times \pi r^2$

Put the value into the formula

$F=2.75\times\pi (0.01)^2$

$F=0.086\ N$

We need to calculate the acceleration

Using formula of force

$F=ma$

Put the value into the formula

$0.086=110\times a$

$a=\dfrac{0.086}{110}$

$a=0.000781\ m/s^2$

$a=7.81\times10^{-4}\ m/s^2$

(a). We need to calculate speed of the block

Using equation of motion

$v^2=u^2+2ad$

Put the value into the formula

$v=\sqrt{2\times7.81\times10^{-4}\times100}$

$v=0.395\ m/s$

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

8 0

2 years ago

Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.

n200080 [17]

Answer:

The answer to your question is the letter A) F = 9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

F = k q₁q₂ /r²

-Substitution

F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

F = 3.512 x 10⁻⁷ / 0.381

-Result

F = 9.227 x 10⁻⁷ N ≈ 9.23 x 10⁻⁷ N

8 0

2 years ago

The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the neares

KengaRu [80]

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, $\theta_1=47.7^{\circ}$

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

$n_1\ sin\theta_1=n_2\ sin(90)$

$n_1\ sin(47.7)=1$

$n_1=\dfrac{1}{sin(47.7)}$

$n_1=1.35$

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.

5 0

2 years ago

A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless t

Molodets [167]

Answer:

Explanation:

a ) angular frequency ω = $\sqrt{\frac{k}{m} }$

k is spring constant and m is mass attached

ω = $\sqrt{\frac{20}{1.5} }$

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

6 0

2 years ago

delivery ladies is Shivam noted at the thundering sound of 6 second after the lightning was seen by him​

delivery ladies is Shivam noted at the thundering sound of 6 second after the lightning was seen by him