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Georgia [21]
3 years ago
15

3. Differentiate between: Law of floatation and Archimedes' principle

Physics
1 answer:
irga5000 [103]3 years ago
4 0

Explanation:

this is the answer I found in my book.

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The work function for magnesium is 3.70 ev. what is its cutoff frequency?
alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

#SPJ1

7 0
2 years ago
In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
Talja [164]

Answer:

η = 40 %  

Explanation:

Given that

Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

\eta=\dfrac{W(net)}{Q(heat\ addition)}

Now by putting the values in the above equation ,then we get

\eta=\dfrac{400}{1000}

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0
3 years ago
You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

f(x,t)=Acos(kx-\omega t)

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

v=\frac{df}{dt}=13.15cos(12.56t)

7 0
3 years ago
What are the two kinds of nucleons?
Akimi4 [234]

Answer:

The main types of nucleons are protons and neutrons. A proton, as its name suggests, has a positive electric charge, and a neutron has a neutral electric charge (meaning that it has no charge). The two in the nucleus of the atom make a positive charge, since the neutron has no charge at all.

Explanation:

5 0
3 years ago
A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro
NARA [144]

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width  = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

5 0
2 years ago
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