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2 years ago

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Atoms can be "cooled" to incredibly low temperatures by letting them interact with a laser beam. Various novel quantum phenomena

appear at these temperatures.
What is the rms speed of cesium atoms that have been cooled to a temperature of 100nK ?

Express your answer to two significant figures and include the appropriate units.

1 answer:

Oksanka [162]2 years ago

8 0

Answer:

the rms speed of cesium atoms that have been cooled to a temperature of 100nK = 0.43cm/s or 0.0043m/s

Explanation:

The concept of root mean square velocity is applied, where the average translational kinetic is related to the actual kinetic energy, the expression for the root mean square is the generated.

The detailed steps and appropriate substitution is as shown in the attachment.

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A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".

Alex787 [66]

Answer:

Part a)

$U = 13 J$

Part b)

$v = 2.28 m/s$

Part c)

$v = 177.66 m/s$

Part d)

$W = 1012.7 J$

Part e)

$v = 2.1 m/s$

Part f)

$E = 1037.2 J$

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

$\theta = 38^o$

so the maximum height reached by the pendulum is given as

$h = L(1 - cos\theta)$

so we will have

$h = L(1 - cos38)$

$h = 1.25(1 - cos38)$

$h = 0.265 m$

now gravitational potential energy of the pendulum is given as

$U = mgh$

$U = 5(9.81)(0.265)$

$U = 13 J$

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(0.265)}$

$v = 2.28 m/s$

Part c)

now by momentum conservation we can say

$mv = (M + m) v_f$

$0.065 v = (5 + 0.065)2.28$

$v = 177.66 m/s$

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

$W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2$

$W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2$

$W = 1012.7 J$

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

$0 = mv_1 + Mv_2$

$0 = 0.065(177.6) + 5.50 v$

$v = 2.1 m/s$

Part f)

Total energy released in the process of shooting of gun

$E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2$

$E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)$

$E = 1037.2 J$

6 0

2 years ago

A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about

Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[ $0^{2} -11^{2}$ ]

x-x₀=6.05* $10^{-2}$ m

=6.05 cm

3 0

1 year ago

Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force o

nevsk [136]

In the circular motion of the hammer, the centripetal force is given by
$F=m \frac{v^2}{r}$
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
$F=m \frac{v^2}{r}=(7.26 kg) \frac{(31.95 m/s)^2}{1.215 m}=6100 N$

4 0

2 years ago

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The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?

Marat540 [252]

Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.

7 0

2 years ago

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The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass

Triss [41]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

spring constant of the spring attached to the input piston, $k=1600\ N.m^{-1}$

mass subjected to the output plunger, $m=40\ kg$

<u>Now, the force due to the mass:</u>

$F=m.g$

$F=40\times 9.8$

$F=392\ N$

<u>Compression in Spring:</u>

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{392}{1600}$

$\Delta x=0.245\ m$

or

$\Delta x=245\ mm$

8 0

2 years ago