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3 years ago

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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. The blast wave traveled at 10° above the horizon. (a) What was the average velocity of the blast wave? (b) Compare this with the speed of sound, which is 343 m/s at sea level.

2 answers:

ziro4ka [17]3 years ago

6 0

Answer:

a) Average velocity = 156.7m/s

b) Speed of blast wave is 0.457 times the speed of sound wave

Explanation:

The step by step calculations is as shown in the attached file.

Neporo4naja [7]3 years ago

3 0

Answer:

Velocity,c = 156.67m/s

Explanation: Altitude, d = 23.5km or 23500m

Time duration, T = 2 minutes + 30 seconds = (120 + 30) seconds = 150 seconds

a) Frequency, F = 1/T= 1/150 Hz

Hence, Velocity, c = d X F = 23500 X 1/150

∴ c = 156.67 m/s

b) The speed of sound increases as the wave moves from a lighter medium to denser medium. While the of the superbolide meteor decreases from a lighter medium to denser medium.

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A straight segment of wire has a length of 25 cm and carries a current of 5A. If the wire is perpendicular to the magnetic field

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Answer:

The magnitude of the magnetic force acting on the conductor is 0.75 Newton

Explanation:

The parameters given in the question are;

The length of the straight segment of wire, L = 25 cm = 0.25 m

The current carried in the wire, I = 5 A

The orientation of the wire with the magnetic field = Perpendicular

The strength of the magnetic field in which the wire is located, B = 0.60 T

The magnetic force, 'F', is given by the following formula;

F = $\underset{I}{\rightarrow }$ ·L× $\underset{B}{\rightarrow }$ = I·L·B·sin(θ)

Where;

$\underset{I}{\rightarrow }$ = The current flowing, I

L = The length of the wire

$\underset{B}{\rightarrow }$ = The magnetic field strength, B

θ = The angle of inclination of the conductor to the magnetic field

Where I = 5 A, L = 0.25 m, B = 0.60 T, and θ = 90°, we get;

F = 5 A × 0.25 m × 0.60 T × sin(90°) = 0.75 N

Therefore

The magnitude of the magnetic force, F = 0.75 N.

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The area of an equilaterla triangle is increasing at a rate of 5 m^2/hr. find the rate at witch the height is chganging when the

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The rate at which the height is changing is ( 5 / x ) m / hr

We know that,

Area of an equilateral triangle A = $\sqrt{3}$ $x^{2}$ / 4

h = $\sqrt{3}$ x / 2

Where,

x = Side

h = Height

Given that,

dA / dt = 5 $m^{2}$ / hr

h = $\sqrt{3}$ x / 2

Differentiate both sides with respect to t

dh / dt = ( $\sqrt{3}$ / 2 ) ( dx / dt )

dx / dt = ( 2 / $\sqrt{3}$ ) ( dh / dt )

A = $\sqrt{3}$ $x^{2}$ / 4

Differentiate both sides with respect to t

dA / dt = ( $\sqrt{3}$ / 4 ) ( 2x ) ( dx / dt )

5 = ( $\sqrt{3}$ / 4 ) ( 2x ) ( 2 / $\sqrt{3}$ ) ( dh / dt )

dh / dt = ( 5 / x ) m / hr

Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt

Therefore, the rate at which the height is changing is ( 5 / x ) m / hr

To know more about Rate of change of height

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A certain pendulum with a 1.00 kg bob has a period of 3.50 s. What will happen to the period of the pendulum if the 1.00 kg bob

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Answer:

The period will be doublef because increasing the mass increases the period linearly.

Explanation:

A stiffer spring with a constant mass decreases the period of oscillation. Increasing the mass increases the period of oscillation

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3 years ago

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the

makkiz [27]

Answer:

(a) She has traveled a total distance of 28958.04 m during the entire trip.

(b) Her average velocity for the trip is 6.12 m/s

Explanation:

Here is the complete question:

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s. During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s. Finally, during the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s. (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

Explanation:

(a) To determine how far the bicyclist has traveled during the entire trip, we will calculate the distance she covered in each part of the trip, and then sum up the distances to determine the total distance covered.

First, The distance covered in the first part of the trip

During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s

That is,

Average speed, $v$ = 7.57 m/s

and time, $t$ = 26.8 minutes

Convert the time to seconds

∴ $t$ = 26.8 minutes = (26.8 × 60) secs = 1608 secs

$Average speed = \frac{Distance }{ Time}$

Then, $Distance = Average speed$ × $Time$

Hence, for the first part

$Distance = 7.57$ × $1608$

$Distance = 12172.56$ m

This is the distance covered in the first part of the trip.

For the distance covered in the second part of the trip

During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s

That is, Average speed, $v$ = 3.17 m/s

and time, $t$ = 42.4 minutes

Convert the time to seconds

∴ $t$ = 42.4 minutes = (42.4 × 60) secs = 2544 secs

From,

$Distance = Average speed$ × $Time$

$Distance = 3.17$ × $2544$

$Distance = 8064.48$ m

This is the distance covered in the second part of the trip.

For the distance covered in the third part of the trip

During the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s

That is, Average speed, $v$ = 15.0 m/s

and time, $t$ = 9.69 minutes

Convert the time to seconds

∴ $t$ = 9.69 minutes = (9.69 × 60) secs = 581.4 secs

From,

$Distance = Average speed$ × $Time$

$Distance = 15.0$ × $581.4$

$Distance = 8721$ m

This is the distance covered in the third part of the trip.

Now for the distance covered during the entire trip,

Total distance = distance covered in the first part of the trip + distance covered in the second part of the trip + distance covered in the third part of the trip

Hence,

Total distance = 12172 m + 8064.48 m + 8721 m

Total distance = 28958.04 m

Hence, she has traveled a total distance of 28958.04 m during the entire trip.

(b) For her average velocity for the trip

Average velocity is given by

$Average velocity = \frac{Total distance traveled}{Total time}$

Total distance traveled = 28958.04 m

Total time = 1608 secs + 2544 secs + 581.4 secs

Total time = 4733.4 secs

Hence,

$Average velocity = \frac{28958.04}{4733.4}$

Average velocity = 6.1178 m/s

Average velocity ≅ 6.12 m/s

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3 years ago

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 66.0 kg , is moving to the right at 2.00

ivann1987 [24]

Answer:

The velocity of skaters after collision is 0.55 m/s and they are moving to the left

Explanation:

Consider m₁ and m₂ are the masses of two skaters and their initial velocity be v₁ and v₂ respectively.

Consider the velocity along right direction as positive while negative for left direction.

According to the problem,

Mass of first skater, m₁ = 66.0 kg

Mass of second skater, m₂ = 75.0 kg

Velocity of first skater, v₁ = + 2.00 m/s

Velocity of second skater, v₂ = -2.80 m/s

Since, the two skaters grab each other after collision. Hence, they are moving with same final velocity v.

Applying conservation of linear momentum,

Momentum before collision = Momentum after collision

m₁v₁ + m₂v₂ = (m₁+m₂)v

Substitute the suitable values in the above equation.

66 x 2 - 75 x 2.8 = (66 + 75 )v

$v= \frac{-78}{141}$

v = -0.55 m/s

The negative sign denotes that both the skaters are moving in the left direction.

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