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3 years ago

Equations to use: v= λ ∙ f v=d/t

Physics

1 answer:

Margarita [4]3 years ago

3 0

b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

$f=\frac{v}{2L}$

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

$v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s$

c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

$v=\frac{d}{t}$

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

$d=vt=(6,000 m/s)(3 s)=18,000 m$

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In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the

yulyashka [42]

Answer:

$\theta=23.7^{\circ}$

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then $M_{\beta_1}=2.2$

Also

$M_{\beta_1}=M_1 sin \beta$ and making $sin\beta$ the subject of the formula then

$sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}$

Making reference to $\theta-\beta-M$ diagram then

$\theta=23.7^{\circ}$

4 0

3 years ago

If a 25 kg lawnmower produces 347 w and does 9514 J of work, for

Igoryamba

Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

-----------------------

Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

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Note: we don't use the mass at all

6 0

3 years ago

(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel

sammy [17]

Answer:

Part a)

$R_1 = 0.072 m$

Part b)

$R_2 = 0.036 m$

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

$R = \frac{mv}{qB}$

now for single ionized we have

$R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}$

$R_1 = 0.072 m$

Part b)

Similarly for doubly ionized ion we will have the same equation

$R = \frac{mv}{qB}$

$R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}$

$R_2 = 0.036 m$

Part c)

The distance between the two particles are half of the loop will be given as

$d = 2(R_1 - R_2)$

$d = 2(0.072 - 0.036)$

$d = 0.072 m$

6 0

3 years ago

Find the heat energy is required to change 2Kg of ice at 0 C to water at 20 C ( specific latent heat of fusion of water = 336000

katrin2010 [14]

We want to find the energy that we need to transform 2kg of ice at 0°C to water at 20°C.

We will find that we must give 840,000 Joules.

First, we must change of phase from ice to water.

We use the specific latent heat of fusion to do this, this quantity tells us the amount of energy that we need to transform 1 kg of ice into water.

So we need 336,000 J of energy to transform 1kg of ice into water, and there are 2kg of ice, then we need twice that amount of energy:

2*336,000 J = 672,000 J

Now we have 2kg of water at 0°C, and we need to increase its temperature to 20°C.

Here we use the specific heat, it tell us the amount of energy that we need to increase the temperature per mass of water by 1°C.

We know that:

specific heat of capacity of water = 4200 J/kg°C

This means that we need to give 4,200 Joules of energy to increase the temperature by 1°C of 1kg of water.

Then to increase 1°C of 2kg of water we need twice that amount:

2*4,200 J = 8,400 J

And that is for 1°C, we need to give that amount 20 times (to increase 20°C) this is:

20*8,400 J = 168,000 J

Then the total amount of energy that we must give is:

E = 672,000 J + 168,000 J = 840,000 J

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3 years ago

What is the distance that light moves in a year called?

goblinko [34]

The distance that light moves in a year is called a light year.
2) Light-year.

6 0

3 years ago

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