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3 years ago

5

Physics

1 answer:

Tanya [424]3 years ago

8 0

Answer:

Reset

Explanation:

Digital methods are the methods that are uses methodological outlook to study societal change and cultural condition of online data. Reset is use to disguise data In digital methods. It is use to set again and conceal data by giving the data a different form. It restores the device to the original manufacture's settings.

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Fynjy0 [20]

C real,inverted and smaller than the object

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3 years ago

Where should a force be applied on a lever arm to produce the most torque

nignag [31]

The answer is,

<u>Farthest from the axis of rotation</u>

Please rate <u>Brainliest</u> (:

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3 years ago

Night vision glasses detect energy given off by objects as (2 points)

Allushta [10]

The answer here would be infrared waves. Hot objects and humans give off heat in the form of infrared light, thermal imaging technology in the goggles enable them to catch this light emitted by these objects

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2 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The

Liula [17]

What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be $\frac{F}{2}$ now:

a= $\frac{F}{2m}$

So the accelaration will also he halved (it's the original acceleratation divided by 2)

b) The object's mass is halved?
a= $\frac{F}{m/2}$ =a= $\frac{F2}{m}$

which is the original acceleration times two!! so it will double

c) The force and the object's mass are both halved?
now we have

a= $\frac{F/2}{m/2}$ =a= $\frac{2F}{2m}$ =a= $\frac{F}{m}$

so they will cancel each other out and the acceleration will stay the same!

5 0

2 years ago