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Iteru [2.4K]
2 years ago
15

A typical meteor that hits the earth's upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at

an impressive 40 km/s. As the meteor slows, the resulting thermal energy makes a glowing streak across the sky, a shooting star. The small mass packs a surprising punch.
At what speed would a 900 kg compact car need to move to have the same kinetic energy?
Physics
1 answer:
attashe74 [19]2 years ago
8 0

Answer:

Answer:u=66.67 m/s

Explanation:

Given

mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg

velocity of meteor v=40km/s \approx 40000 m/s

Kinetic Energy of Meteor

K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}

K.E.=2\times 10^6 J

Kinetic Energy of Car

=\frac{1}{2}\times Mu^2

=\frac{1}{2}\times 900\times u^2

\frac{1}{2}\times 900\times u^2=2\times 10^6  

900\times u^2=4\times 10^6

u^2=\frac{4}{9}\times 10^4

u=\frac{2}{3}\times 10^2

u=66.67 m/s

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If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100
SCORPION-xisa [38]

Answer:

Charge enter a 0.100 mm length of the axon is 8.98\times 10^{-12} C

Explanation:

Electric field E at a point due to a point charge is given by

E=k \frac{q}{r^2}

where k is the constant =9.0 \times 10^9  Nm^2 / C^2

q is the magnitude of point charge and r is the distance from the point charge

Charges entering one meter of axon is 5.\times 10^{11} \times (+e)

Charges entering 0.100 mm of axon is 5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}

substituting the value of +e=1.6\times 10^{-19} C in above equation, we get charge enter a 0.100 mm length of the axon is

q=5.\times 10^{11} \times1.6\times 10^{-19}  \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C

3 0
2 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
2 years ago
The starter armature is rubbing on the field coils. technician a says the bushings need to be replaced. technician b says the br
Viefleur [7K]
In a situation in which the smarter armiture is rubbing on the field the most commin reason are defective bushings. So, the ststement the techician A ssys that the brushings shojld be replaced is correct. Techinican A is right.
6 0
2 years ago
Read 2 more answers
Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o
mel-nik [20]

Answer:

a. 3/4λ

d. 1/4λ

Explanation:

When the wavelength of the sound waves is λ and the two waves are having same frequency the waves are said to be out of phase if their phase difference is in the multiples of \frac{\lambda}{2} or 180°.

When the two waves are out of phase then their opposite maxima coincide at the same time resulting in the minimum amplitude of the resulting wave throughout.

  • As we observe from the schematic that the a wave has sinusoidal pattern of variation and we get a maxima after each \frac{\lambda}{4} of the distance.
  • Here we have two speakers out of phase therefore on shifting one of the speakers by the odd multiples of \frac{\lambda}{2} we have the maxima or the extreme amplitudes.

So, we must place the microphone at  3/4λ and 1/4λ to pickup the loudest sound.

4 0
2 years ago
An object moving at a velocity of 32m/s slows to a stop in 4 seconds. What was its acceleration?
Romashka [77]

Answer:

8m/s

Explanation:

a=d/t

a=32/4

a=8 m/s

6 0
2 years ago
Read 2 more answers
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