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madreJ [45]
3 years ago
7

Collisions between atoms are often elastic, but sometimes inelastic collisions occur, and the lost kinetic energy can become int

ernal energy in the atoms, exciting electrons to higher energy orbitals or even ejecting then from the atom entirely, which is called ionization (and the required energy the ionization energy).2 (a) A Cs 133 atom (mass 133 amu) has an ionization energy of 4.0 eV.3 Express this energy in joules. (b) What is the minimum speed that a O16 atom can have if it is to ionize a cesium atom at rest upon impact?
Physics
1 answer:
Galina-37 [17]3 years ago
4 0

Answer:

a) E = 6.4 1019 J    b)  v = 0.69 10⁴4 m / s

Explication

a) convert E = 4.0 eV

    1 eV = 1.6 10⁻¹⁹ J

   E = 4.0 eV (1.6 10⁻¹⁹ J / 1 eV)

   E = 6.4 10⁻¹⁹ J

b) Suppose we have a frontal shock and all the kinetic energy of oxygen is transferred to Cs

    Ei = K = ½ m v²

    Ef = 6.4 10⁻¹⁹ J

    ½ m v² = 6.4 10⁻¹⁹

The oxygen mass of the periodic table is

     PA = 15,999 u

     1u = 1.660 10⁻²⁷ kg

     Pa = 15,999 1,660 10⁻²⁷ kg

     m= Pa = 26,558 10⁻²⁷ kg

Let's calculate the speed

    v2 = 2 / m 6.4 10⁻¹⁹

    v2 = 2 / 26,558 10⁻²⁷ 6.4 10⁻¹⁹ =

    v = √0.4819 10⁸

    v = 0.69 10⁴4 m / s

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A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50
PSYCHO15rus [73]

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

T=2\pi\sqrt{\dfrac{I}{mgh}}

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

I=I_{cm}+md^2

For a meter stick mass m , the rotational inertia about it's center of mass

I_{cm}-\dfrac{mL^2}{12}

Where, L = 1 m

Put the value into the formula of time period

T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}

T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}

T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0

Put the value of T, L and g into the formula

4.028d^2-6.25d+0.336=0

d = 0.056\ m, 1.496\ m

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

T=2\pi\sqrt{\dfrac{l}{g}}

Put the value into the formula

T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}

T=1.35\ sec

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

7 0
3 years ago
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou
zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}

E=6.88\times10^{7}\ N/C

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}

E=5.43\times10^{6}\ N/C

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0
3 years ago
Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal
Arturiano [62]

Answer:

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² =  5880000 / 0.018

ω ² =  326666667

ω = 18074 rad/sec

7 0
3 years ago
in this model, the velocity of the spacecraft at position 2 is A.) equal to B.) greater than C.) less than the velocity of the c
finlep [7]
<h2>1. Right answer: the velocity of the spacecraft at position 2 is <u>greater than</u> the velocity of the craft at position 4. </h2>

This is due the gravity field of the planet (The Earth in this case) is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft  to enter a circular orbit.

<h2>2. Right answer: At position 1, the direction of the spacecraft changes because of <u>the gravitational force between Earth and the spacecraft. </u></h2>

As explained in the prior answer, this is the exact and correct point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.


<h2>3. Right answer: Position 3 represents <u>the orbital path or velocity of Earth </u></h2>

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished.

If the orbital path of the Earth were the opposite from the shown in the figure, the effect on the craft would be braking.

Note all of these is related to the <u>gravitational assistance. </u>

<u>Gravitational assistance</u> is the maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe changing its trajectory.

This maneuver is also called <em>slingshot effect, swing-by</em> or <em>gravity assist</em>. It is a common technique in space for the outer Solar System missions , in order to save costs in the launch rocket and thrusters.


6 0
3 years ago
Read 2 more answers
A coach is hitting pop flies to the outfielders. if the baseball (m= 145 g) stays in contact with the bat for 0.04 s and leaves
vekshin1
At the start, the ball is at rest and therefore, u=0 m/s. As it leaves the bat, v= 50 m/s

From equations of motion, v=u+at = at (since u=o)
a=v/t = 50/0.04 = 121250 m/s^2

From Newton's second law,
F=ma = 145/1000 *1250 = 181.25 N
4 0
3 years ago
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