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Setler [38]
3 years ago
6

A 1400-kg car moving with a speed of 32 m/s is approaching a stoplight. The light turns yellow and the car makes an abrupt stop

over a distance of 60 meters. The amount of work done on the car approximately.
a. 84,00 J
b. 268, 800 J
c. 716, 800 J
d. 840,000 J
e. 2, 688,000 J
f. None of these are correct.
Physics
1 answer:
inna [77]3 years ago
8 0

Answer:

c. 716, 800 J

Explanation:

t = Time taken

u = Initial velocity = 32 m/s

v = Final velocity = 0

s = Displacement = 60 m

a = Acceleration

m = Mass of car = 1400 kg

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-32^2}{2\times 60}

Work done is given by

W=Fscos\theta\\\Rightarrow W=mascos\theta\\\Rightarrow W=1400\times \dfrac{0^2-32^2}{2\times 60}\times 60\times cos0\\\Rightarrow W=-716800\ J

The amount of work done to stop the car is 716800 J

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Your answer is C) The speed of sound is higher in solids than in liquids. 
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2 years ago
An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic
shepuryov [24]

Answer:

v=\sqrt{\frac{kZe^2}{mr}}

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2} (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

F=m\frac{v^2}{r} (2)

where

m is the mass of the electron

v is the speed of the electron

Combining the two equations (1) and (2), we find

k\frac{Ze^2}{r^2}=m\frac{v^2}{r}

And solving for v, we find an expression for the speed of the electron:

v=\sqrt{\frac{kZe^2}{mr}}

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2 years ago
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igomit [66]

the answer is the forth one treatment of cancer

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Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Burka [1]

Answer:

1.91773\times 10^{37}\ kg

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = \dfrac{d}{2}=\dfrac{16}{2}=8\ ly

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

1\ ly=9.461\times 10^{15}\ m

As the centripetal force balances the gravitational energy we have the following relation

\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg

Mass of the the massive object at the center of the Milky Way galaxy is 1.91773\times 10^{37}\ kg

4 0
3 years ago
Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the
xeze [42]
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is 
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
6 0
3 years ago
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