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3 years ago

Help me with number 3 and 4 pleaseeeeeeeee

Physics

2 answers:

Andrej [43]3 years ago

6 0

i think its b cause if you were talking about ice it freezes the melts the evaporates to gas

Roman55 [17]3 years ago

3 0

Number three I think might be b

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Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc

kaheart [24]

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

$f'=f_{1}-f$ ...(I)

$f'=f_{1}+f$ ...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

$f'=462-5=457\ Hz$

$f'=462+5=467\ Hz$

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

4 0

3 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

$F =326.7 \ N$

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

3 years ago

Light travels in a straight line at a constant speed of 300000 m/s. What is the acceleration of light?

mafiozo [28]

Answer:

As light travels in a straight line at a constant speed, it's acceleration is <u>0 m/s²</u>.

There is no rate of change of speed, so there is no acceleration.

<u>0 m/s²</u> is the right answer.

4 0

3 years ago

8. What is the frequency of green light waves that have a wavelength of 5.2 x 10-7 m.? The speed of light is 3.0 x 108 m/s

o-na [289]

Answer:

$f=5.76\times 10^{14}\ Hz$

Explanation:

We need to find the frequency of green light having wavelength o $5.2\times 10^{-7}\ m$ . It can be calculated as follows :

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{5.2\times 10^{-7}}\\\\f=5.76\times 10^{14}\ Hz$

So, the required frequency of green light is equal to $5.76\times 10^{14}\ Hz$ .

4 0

2 years ago

A 10,000-watt radio station transmits at 535 kHz. Determine the number of joules transmitted per second. 10,000 J/s 10 J/s 535 J

REY [17]

1 nanowatt = 1 nanojoule/sec
1 watt = 1 joule/sec
10 watts = 10 joules/sec
100 watts = 100 joules/sec
742.914 watts = 742.914 joules/sec
1,000 watts = 1,000 joules/sec
10,000 watts = 10,000 joules/sec
100,000 watts = 100,000 joules/sec
1 megawatt = 1 megajoule/sec
1 gigawatt = 1 gigajoule/sec
1 petawatt = 1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

5 0

3 years ago