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Damm [24]
3 years ago
13

On April 13, 2029 (Friday the 13th!), the asteroid 99942 mi Apophis will pass within 18600 mi of the earth-about 1/13 the distan

ce to the moon! It has a density of 2600 kg/m^3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.
1)If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver?

2)The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184x10^15 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
Physics
1 answer:
Sergeu [11.5K]3 years ago
4 0

Answer:

Explanation:

Volume of asteroid = 4/3 x π x 160³

= 17.15 x 10⁶

mass = volume x density

= 17.15 x 10⁶ x 2600

= 445.9 x 10⁸ kg

kinetic energy

= 1/2 x 445.9 x 10⁸  x( 12.6 )² x 10⁶

= 35.4 x 10¹⁷ J .

2 )

energy of 15 megaton

= 4.184 x 10¹⁵ x 15 J

= 62.76 x 10¹⁵ J

No of bombs required

= 35.4 x 10¹⁷ / 62.76 x 10¹⁵

= 56.4 Bombs .

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This is a binary star system
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4 years ago
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Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
Find the current i in:
lawyer [7]

Answer:

I = 24 A

Explanation:

This is Parallel Circuit and it is the first principle of parallel circuit that voltage will be equal in all components in the circuit

It includes 10 resistors Therefore the voltage across,

R1 = R2 = R3 = R4 = R5 = R6 = R7 = R8 = R9 = R10 = voltage in battery

<h2>Ohm's Law</h2>

We will apply Ohm's Law to each resistor to find its current because we know the voltage across each resistor is 12 V  and the resistance of each resistor is 5Ω

I (R1) = E (R1) / R1

I (R1) = 12v / 5Ω

I (R1) = 2.4 A

The value resistance E of all resistors are same therefore by applying the formula above the value of current in all resistors will be 2.4 A

The Total current in the circuit will be

I (total) = I (1) + I (2) + I (3) + I (4) + I (5) + I (6) + I (7) + I (8) + I (9) + I (10)

I (total) = 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4

I (total) = 24 A

5 0
3 years ago
An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = \frac{2L}{3}

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center  = 0.25 × wavelength

Distance = 0.25 x \frac{2L}{3}

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = 0.25 x \frac{2 \times 39}{3}

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

7 0
3 years ago
What kind of pollution causes nitrogen and phosphorous to enter bodies of water?
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Nutrient pollution. I just took the quiz and got it right
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