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3 years ago

How can you tell the difference between an expression of speed and an expression of velocity?

1 answer:

6 0

Answer= D. Velocity includes a direction; speed does not.

Velocity and speed both have the same units of m/s, so answers A and B are wrong.

Velocity is a vector quantity and so has direction and magnitude, so C is wrong.

Speed is a scalar quantity, only having magnitude and no direction.

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The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric fi

e-lub [12.9K]

Answer:

please the answer below

Explanation:

A general electromagnetic plane wave, traveling in the x direction, can be expressed in the form:

$\vec{E}=E_0e^{-i(k\cdot x-\omega t)}\hat{j}\\\\\vec{B}=B_0e^{-i(k\cdot x-\omega t)}\hat{k}\\\\$ (1)

a. amplitudes

from (1) we can observe that E_0 and B_0 are the amplitudes.

2. frequency

By replacing (1) we obtain:

$\vec{E}=E_0e^{-i(k(0)-\omega t)}\hat{j}=E_0[cos\omega t+sin\omega t]\hat{j}$

the wave respond to the followinf physical variables: amplitude, frequency, time and position, as we can see in (1).

hope this helps!!

7 0

3 years ago

You place a mass of 250 grams on the measurement tray of a triple beam balance and then set the rider on the 500 gram beam to th

WARRIOR [948]

"<span>The pointer will be above the zero mark" is the one statement among the following choices given in the question that is true. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your great help.</span>

3 0

4 years ago

Read 2 more answers

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

4 years ago

Se pune<br>Numerical problems<br>Convert 100ºC to Kelvin scale

IgorC [24]

Answer:

373 K

Explanation:

To convert from °C to kelvin, you add the value to 273

Therefore,

100°C = (100+273)K

= 373K

5 0

4 years ago

What do you mean by reflection of sound

Bogdan [553]

Answer:

Reflection of sound waves also leads to echoes. Echoes are different than reverberations. Echoes occur when a reflected sound wave reaches the ear more than 0.1 seconds after the original sound wave was heard. ... Reflection of sound waves off of curved surfaces leads to a more interesting phenomenon.

8 0

3 years ago