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3 years ago

If a seismic wave has a period of 0.0202s, find the frequency of the wave.

Physics

1 answer:

maw [93]3 years ago

7 0

Answer:

49.5 Hz.

Explanation:

From the question given above, the following data were obtained:

Period (T) = 0.0202 s

Frequency (f) =?

The frequency and period of a wave are related according to the following equation:

Frequency (f) = 1 / period (T)

f = 1/T

With the above formula, we can obtain the frequency of the wave as follow:

Period (T) = 0.0202 s

Frequency (f) =?

f = 1/T

f = 1/0.0202

f = 49.5 Hz

Therefore the frequency of the wave is 49.5 Hz.

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Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

$P_2 = 44.15Lbf/in^2$

Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

5 0

3 years ago

The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$

7 0

3 years ago

An electron has ___________

Sauron [17]

Answer:

An electron has a negative electric charge.

3 0

3 years ago

Read 2 more answers

8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?

Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

$q_1=q_2=+1C$ are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

$F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N$

8 0

3 years ago

What type of fat may help lower the risk of heart disease

enyata [817]

Polyunsaturated fatty acids which is Omega-3 fatty acids.

8 0

3 years ago