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3 years ago

State the name of the process in each of the following changes of state:

Physics

1 answer:

polet [3.4K]3 years ago

4 0

Answer:

A. Melting

B. Freezing

C. Condensing

D. Boiling

Explanation:

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A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million

hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0

3 years ago

Read 2 more answers

Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips

Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

$\frac{f'}{f} = \frac{v'}{v}$

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

4 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

3 years ago

Why are there so few working wave power stations?

Bond [772]

Well mostly because it's for oceans and lakes that have waves. so if you don't live near any of these, you wouldn't be getting power.

3 0

3 years ago

Read 2 more answers

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago