All categories

2 years ago

On a 30degrees day, there is an explosion. The sound is heard 3.4s after seeing the flash. How far away was the explosion

Physics

1 answer:

chubhunter [2.5K]2 years ago

5 0

Answer:

Distance = 1.2 km (Approx)

Explanation:

Given:

Temperature (T) = 30°C

Total Time taken (t) = 3.4 Sec

We know that sound increases in the air sound increase nearly 0.60 m/s for every sound with temperature.

Speed of sound = 331 m/s

So,

V = (331 + 0.60T) m/s

V = 331+(0.60×300C) m/s

V = 349 m/s

Distance = speed × time

Distance = 349 × 3.4

Distance = 1,186.6 m

Distance = 1.2 km (Approx)

You might be interested in

What are examples of government action? Please HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

alexira [117]

The answer is <span>Encourage private ownership Profit motives and promote competition and monopoly regulate companies sell food and regulate pollution</span>

6 0

2 years ago

Define couple and give 2 examples

Elodia [21]

Answer:

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation.

For example, turning a key in a lock and turning a steering wheel.

3 0

2 years ago

Read 2 more answers

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

2 years ago

If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now

Andru [333]

Answer:

The new period will be √6 *T

Explanation:

period ,T=2π√(L/g) ................equation 1

where T is the period on earth

gravitational acceleration on the moon is g/6

T1 = 2π√[L/(g/6)]

T1=2π√(6L/g) ...............equation 2

divide equation 2 by 1

T1/T =2π√(6L/g)÷2π√(L/g)

T1/T =√(6L/L)

T1/T =√6

T1 = √6 *T

5 0

2 years ago

A round object of mass 10 kg and radius 0.5 m rolls without slipping down a hill from a height of 4.5 m. If its speed at the bot

Mice21 [21]

Answer:

moment of inertia is 2.72 kg m²

Explanation:

given data

mass m = 10kg

height h = 4.5 m

radius r = 0.5 m

speed v = 6.5 m/s

to find out

moment of inertia

solution

we apply here conservation of energy

that is

mgh = 1/2 ×mv² + 1/2 × Iω²

here I is moment of inertia we find and

we know ω = Velocity / radius = 6.5 / 0.5 = 13

and g = 9.8

so put here all these value

10 (9.8) 4.5 = 1/2 ×(10)(6.5)² + 1/2 × I(13)²

441 = 211.25 + 1/2 × I( 169 )

I = 2.72

so moment of inertia is 2.72 kg m²

7 0

2 years ago