What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?

Answers

Answer 1

Answer: Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
= (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
= (9.1×10∧-31 × 1,153,846.154∧2) ÷2
= (1.21154×10∧-18) ÷ 2
= 6.05769×10∧-19 J

Answer 2

Answer:

Answer:

K = 6.02 × 10⁻¹⁹ J

Explanation:

The momentum (p) of an electron is its mass (m) times its speed (v).

p = m × v

v = p / m = (1.05 × 10⁻²⁴ kg.m/s) / 9.11 × 10⁻³¹ kg = 1.15 × 10⁶ m/s

We can find the kinetic energy (K) using the following expression.

K = 1/2 × m × v²

K = 1/2 × 9.11 × 10⁻³¹ kg × (1.15 × 10⁶ m/s)²

K = 6.02 × 10⁻¹⁹ J

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An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materials have this space/size? (d) Assuming that we can measure the velocity to an accuracy of 10%. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position.

Answers

Answer:

P = 2.91*10^{-24} kg m/s

$\lambda = 2.73 *10^(-10) m$

size of atom hat lie in range of 1 to 5 Angstrom

$\Delta x = 0.2272$ Angstrom

Explanation:

A) MOMENTUM

p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

$\lambda = (h)/(mv)$

where h is plank constant

so $\lambda = (6.626*10^(-34))/(2.91*10^(-24))$

$\lambda = 2.73 *10^(-10) m$

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

$(\Delta v)/(v) = 0.1$

$\Delta v = 0.1 v$

we know that

$\Delta p *\Delta x = (h)/(4\pi)$

$m \Delta v \Delta x =(h)/(4\pi)$

$\Delta x = (h)/(m \Delta v)$

$\Delta x = (2.272)/(0.1)$ [ $\Delta v = 0.1 v$ ]

$\Delta x = 0.2272$ Angstrom

The thickness of a $1 bill is 0.11 mm. If you have a stack of $1 bills 450 m tall, how much money do you have?

Answers

Answer:

4090909

Explanation:

Thickness of one bill = 0.11 mm

Total thickness = 450 m

No of $1 bills = total thickness / thickness of one bill

No of $1 bills = 450 / 0.11 × 10^-3

= 4090909

Answer:

You will have 4.5 million dollar

Explanation:

The thickness of a $1 bill is 0.11 mm

So we have

1 $ = 0.1 mm

0.1 mm = 1 $

0.0001 m = 1 $

1 m = 10000 $

450 m = 450 x 10000 = 4500000 $

So you will have 4.5 million dollar

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?

Answers

Answer:

$(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = (d)/(2)$

$r = (0.03)/(2)$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^(-4 ) \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o( d \phi )/(dt )$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

$\mu_o I + \epsilon_o \mu _o( d \phi )/(dt ) = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^(-12 ) \ m^(-3) \cdot kg^(-1)\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

$\mu_o I + \epsilon_o \mu _o ( d [EA] )/(dt ) = 0$

=> $(dE)/(dt) =- (I)/(\epsilon_o * A )$

=> $(dE)/(dt) =- (17)/(8.85*10^(-12) * 7.07*10^(-4) )$

=> $(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

Dogs can hear higher-pitched whistles that humans do. How do you
think the sound frequencies that dogs can
hear compare to the frequencies that humans
can hear?

Answers

Dogs can hear sounds at higher frequencies than humans. The range of sound frequencies that dogs can hear is approximately 40 Hz to 60,000 Hz, while the range for humans is 20 Hz to 20,000 Hz. This means that dogs can hear ultrasonic sounds that are beyond the range of human hearing.

What is sound about?

In terms of physics, sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave.

Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Dogs have the ability to hear ultrasonic sounds that are audible only to them.

Learn more about sound on:

brainly.com/question/16093793

#SPJ1

A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire