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aleksandrvk [35]
3 years ago
13

On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth's atmosphere over Chelyabinsk, Russia, and expl

oded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity (in m/s) of the blast wave? (Enter the magnitude.) x m/s (b) Compare this with the speed of sound, which is 343 m/s at sea level. Volast wave = sound
Physics
1 answer:
fredd [130]3 years ago
7 0

Answer:

156.67 m/s

0.45676 times the speed of sound

Explanation:

Distance from the ground = 23.5 km = 23500 m

Time taken by the blast waves to reach the ground = 2\ minutes\ 30\ seconds=2\times 60+30=150\ s

Spedd of the wave would be

Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s

The velocity of the blast wave is 156.67 m/s

v = Velocity of sound = 343 m/s

\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v

The blast wave is 0.45676 times the speed of sound

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Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

6 0
3 years ago
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition
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Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

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Here A = 13 and B = 427

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As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

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K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m

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\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}

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3 years ago
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Answer:

f = 8 %

Explanation:

given,

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density of water = 1000 kg/m²

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1080 + f x 1.2 = 1000 + 1000 f      

998.8 f = 80                                  

f = 0.0800                            

f = 8 %                                        

the volume increase factor of fish will be equal to f = 8 %

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