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MaRussiya [10]
3 years ago
13

What orientation of velocity and acceleration will cause something to initially speed up?

Physics
1 answer:
Temka [501]3 years ago
7 0
-- If acceleration and velocity are in the same direction,
then the object is speeding up.

-- If acceleration and velocity are in opposite directions,
then the object is slowing down.

-- If acceleration is perpendicular to velocity, then the object
is moving on a circular curve at constant speed.
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M = 1250 kg
V = 14.0 m/s
t = 7.25 s
F = ?

Impulse equation
F . t = m . V

:. F = mv/t

F = (1250 kg x 14.0 m/s) / 7.25s

F = 2414 N
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The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto
igor_vitrenko [27]

a) The angular velocity of the electron is 4.12\cdot 10^{16} rad/s

b) The number of revolutions per second is 6.54\cdot 10^{15} rev/s

c) The centripetal acceleration of the electron is 8.98\cdot 10^{22} m/s^2

Explanation:

a)

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

\omega = \frac{v}{r}

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the angular velocity is

\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s

b)

The period of revolution of the electron is given by

T=\frac{2\pi}{\omega}

where

\omega = 4.12\cdot 10^{16}rad/s is the angular velocity

Substituting,

T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15} rev/s

c)

The centripetal acceleration of an object in circular motion is

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circle

For the electron, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the centripetal acceleration is

a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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4 years ago
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