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2 years ago

Find the frequency of a wave with a period of 0.5 seconds

Physics

1 answer:

strojnjashka [21]2 years ago

8 0

Frequency of the wave is 2 per second

Explanation:

Frequency is the number of times waves pass at a particular point of time. Here, time period = 0.5 s

Frequency is given by the formula

f = 1/T, where f is the frequency and T is the time period

⇒ f = 1/0.5 = 2 per second

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Can someone answer and explain these questions?

Andru [333]

Answer:

oh that looks hard but sorry im not good at math ether so sorry i couldnt help

Explanation:

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2 years ago

4. A hiker leaves camp and using a compass walks 4 km East, 6 km South, 3 km East, 5 km North, 10 km West, 8 km North and 3 km S

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Answer:

Because the hiker walked directly west and then directly north the two legs of the hike forms a right triangle. Therefore we can use the Pythagorean Theorem to solve this problem.

c=5

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2 years ago

Which one of the following statements about sulfric acid is correct?

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Answer:

Which one of the following statements about sulphuric acid is correct?

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3 years ago

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

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3 years ago

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