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Tatiana [17]
3 years ago
5

Suppose a hydrogen atom in its ground state moves 130 cm through and perpendicular to a vertical magnetic field that has a magne

tic field gradient dB/dz = 1.2 × 102 T/m. (a) What is the magnitude of the force exerted by the field gradient on the atom due to the magnetic moment of the atom's electron, which we take to be 1 Bohr magneton? (b) What is the vertical displacement of the atom in the 130 cm of travel if its speed is 2.2 × 105 m/s?

Physics
1 answer:
blagie [28]3 years ago
5 0

Answer:a)1.11×10^-21Nm

b) 1.16×10^-3m

Explanation:see attachment

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

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

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
Sometimes in the winter, ice crystals grow directly from the moisture in the air. As the water changes from a gas to a solid, th
mrs_skeptik [129]

Answer: B. slow down as they lose energy

7 0
3 years ago
Explain how an intrusive igneous rock could become a metamorphic rock and then an extrusive igneous rock.
sleet_krkn [62]
1.igneous rocks are formed when magma or lava cools and hardens
2. the cooling rate of the rock
5 0
3 years ago
The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s
Anastasy [175]
Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now
6 0
3 years ago
If a bowling ball of mass 7.0 kg has a potential energy of 1100 Joules, how high above the ground is the bowling ball?
kirza4 [7]

Answer:

16.03m(2dp)

Explanation:

Ep=m x g x h

1100=7.0x 9.8( gravitational field strength) x h

Height= 1100/7.0 x 9.8

          =16.03498542

          = 16.03m (2dp)

6 0
3 years ago
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