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IrinaK [193]
3 years ago
7

A wave has a frequency of 875 hz and a wavelength of 352 m. At what speed is this wave traveling 

Physics
1 answer:
Sonbull [250]3 years ago
5 0

Answer:

308,000 or 30.8×10^3

Explanation:

v=f×lamda

v is ?, f is 875Hz, lamda is 352m

v=875×352

v=308,000

v=30.8×10^3 m/s

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HELPPPP MEEE!!! 15 POINTS
GalinKa [24]

Answer:

I think it is eyeglasses

Explanation:

I think this because the glasses usually reflect the same thing sometimes and the light passes through really easy

4 0
4 years ago
an object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second. the object strikes the ground 3.
Olegator [25]
You know that the formula for finding horizontal displacement is; 

s=uₓt 
s=(5 m/s)(3s) 
s=15 m

Therefore, the ball travelled 15 m. 

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7 0
3 years ago
Read 2 more answers
What is the total resistance of a 6 ohm and a 8 ohm resistor connected in series?<br> (1 Point)
Oksanka [162]

Answer:

Total Resistance in circuit is Fourteen Ohms <u>(14 Ω).</u>

<u></u>

Explanation:

How do we know, if the resistors are connected in series, the total resistance is the <u>sum of all the resistors.</u>

(Important: The total resistance can only be added just when the resistors are <u>connected in series</u>)

Then, total resistance (<em>TR </em>) is the sum of all resistors (<em>T1 + T2</em>, in this case)

TR = T1 + T2

According to problem data, we have:

TR = 8 Ω + 6 Ω

TR = 14 Ω

                                                                              ║Sincerely, ChizuruChan║

4 0
3 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

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

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
The change from liquid water to water vapor is called___
Jet001 [13]
The answer is a.evaporation

hope this helps!

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7 0
3 years ago
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