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3 years ago

Certain x-rays have a frequency of 1.0×1019hz. calculate their wavelength in air.

1 answer:

6 0

For any electromagnetic wave, the relationship between frequency and wavelength is given by
$\lambda = \frac{c}{f}$
where
$\lambda$ is the wavelength
c is the speed of light (the speed of the electromagnetic wave)
f is the frequency

For the X-rays in our problem, the frequency is $f=1.0 \cdot 10^{19}Hz$ , while the speed of light is $c=3.0 \cdot 10^8 m/s$ , so the wavelength of this radiation is
$\lambda= \frac{3 \cdot 10^8 m/s}{1.0 \cdot 10^{19} Hz} =3 \cdot 10^{-11} m$

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Answer:

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4 years ago

Which of the following statements is/are true? Check all that apply. A. A potential energy function can be specified for a conse

Otrada [13]

Answer:

A. A potential energy function can be specified for a conservative force.

TRUE

because there is no loss of energy in conservative type of forces.

B. A nonconservative force permits a two-way conversion between kinetic and potential energies.

FALSE

It is not true because energy is not conserved in non-conservative forces.

C. The work done by a nonconservative force depends on the path taken.

TRUE

It depends on total path length while in conservative it only depends on initial and final state

D. A potential energy function can be specified for a nonconservative force.

FALSE

Since energy is not stored in non-conservative forces so it is not defined for non conservative forces

E. A conservative force permits a two-way conversion between kinetic and potential energies.

TRUE

Work done against conservative forces is stored in form of potential energy so it is possible to have two way conversion.

F. The work done by a conservative force depends on the path taken.

FALSE

Conservative force work depends only on initial and final state

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4 years ago

A 20-kg child is coasting at 3.3 m/s over flat ground in a 4.0-kg wagon. The child drops a 1.0-kg ball out the back of the wagon

Shalnov [3]

Answer:

The final velocity is $u_f = 3.44 \ m/s$

Explanation:

From the question we are told that

The mass of the child is $m_1 = 20 \ kg$

The initial speed of the child is $u_1 = 3.3 \ m/s$

The mass of the wagon is $m_w = 4.0 \ kg$

The initial speed of the wagon is $u_w = 3.3 \ m/s$

The mass of the ball is $m_2 = 1.0 \ kg$

The initial speed off the ball is $u_2 = 3.3 \ m/s$

Generally the initial speed of the system (i.e the child , wagon , ball) is

$u_1 = u_w = u_2 = u =3.3 \ m/s$

Generally from the law of linear momentum conservation

$p_i = p_f$

Here $p_i$ is the momentum of the system before the ball is dropped which is mathematically represented as

$p_i = ( m_1 + m_2 + m_3 ) * u$

=> $p_i = ( 20 + 4 + 1 ) * 3.3$

=> $p_i = 82.5 \ kg \cdot m/s$

and

$p_f$ is the momentum of the system after the ball is dropped which is mathematically represented as

$p_f = ( m_1 + m_w ) * u_f$

=> $p_i = ( 20 + 4 ) * u_f$

$82.5 = 24 * u_f$

=> $u_f = 3.44 \ m/s$

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3 years ago

The membrane of the axon of a nerve cell is a thin cylindrical shell of radius r = 10-5 m, length L = 0.32 m, and thickness d =

maksim [4K]

Answer:

5.3 x 10⁻⁹ C

Explanation:

r = radius of cylindrical shell = 10⁻⁵ m

L = length = 0.32 m

A = area

Area is given as

A = 2πrL

A = 2 (3.14) (10⁻⁵) (0.32)

A = 20.096 x 10⁻⁶ m²

d = separation = 10⁻⁸ m

$k_{appa}$ = dielectric constant = 4

Capacitance is given as

$Q=\frac{k_{appa}\epsilon _{o}A}{d}$ eq-1

V = Potential difference across the membrane = 74 mV = 0.074 Volts

Q = magnitude of charge on each side

Magnitude of charge on each side is given as

Q = CV

using eq-1

$Q=\frac{k_{appa} \epsilon _{o}AV}{d}$

Inserting the values

$Q=\frac{4 (8.85\times 10^{-12})(20.096\times 10^{-6})(0.074)}{10^{-8}}$

Q = 5.3 x 10⁻⁹ C

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3 years ago