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3 years ago

Which statements describe how a machine can help make work easier? Check all that apply.

1 answer:

7 0

I don't see any answers, but I do know that machines help for example: make chocolate bars. people do not have to cut the bars or another example: salt or season chips. or washing laundry. or electric cutters help so we do not have to do it by hand or with a knife.
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Which steps are important when designing and conducting a scientific experiment

artcher [175]

Having your space clean. have on close toed shoes. have your hair pulled back into a ponytail. keep ur work space clean. wear gloves and goggles. do not have on droopy clothes. follow the steps on the board and double check them.

8 0

4 years ago

(I NEED THIS ANSWERED NOW PLEASE)

VMariaS [17]

Answer:

Zeros that follow non-zero numbers and are also to the right of a decimal point are significant.

Explanation:

For example:

0.300 has 3 significant figures.

5.400 has 4 significant figures.

4 0

2 years ago

Which of the following is not an example of work?

barxatty [35]

<span>The following which is not an example of work is </span>C. holding a tray in the cafeteria line because <span>if force displaces an object it should work. I think it's clear and I am pretty sure this answer will help you.</span>

8 0

3 years ago

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

A trumpet creates a sound wave that has a wave speed of 350m/s and a wavelength of 0.8 m . What is the frequency of the sound wa

ValentinkaMS [17]

Answer:

The frequency of sound wave created by trumpet is 437.5Hz

Explanation:

Given

the speed of sound wave = 350 m $s^{-1}$

the wavelength of sound wave = 0.8 m

the frequency of sound wave = ?

All the waves have same relationship among wavelength, frequency and speed, which is given by the equation:

v = fλ, where

v is speed of the wave

f is frequency of the wave

λ is wavelength of the wave

therefore frequency of sound wave is given by

f = v/λ

= 350m $s^{-1}$ /0.8m

= 437.5 $s^{-1}$

= 437.5Hz (since 1 $s^{-1}$ = 1 Hz (Hertz)

Hence the frequency of sound wave created by trumpet is 437.5Hz

7 0

3 years ago