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3 years ago

9

The period of a mechanical wave is 4 seconds. What is the frequency of the wave?

2 answers:

Artemon [7]3 years ago

8 0

It is 0.25 waves second

kvasek [131]3 years ago

8 0

Answer: D

d. 2.50

Minds On Physics the App ("MOP the App") is a series of interactive questioning modules for the student that is serious about improving their conceptual understanding of physics. Each module of the series covers a different topic and is further broken down into sub-topics. A "MOP experience" will provide a learner with challenging questions, feedback, and question-specific help in the context of a game-like environment. It is available for phones, tablets, Chromebooks, and Macintosh computers. It's a perfect resource for those wishing to refine their conceptual reasoning abilities. Part 5 of the series includes topics on Wave Motion.

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Newton began his academic career in 1667. For how long was he a working scientist? Was he a very productive scientist?

rjkz [21]

Newton's 3 laws are...

inertia: things tend to continue to do what they are doing.

Change: to make something change you need a force to change it. the force needed = the mass times its acceleration

<span> Resistance: When you push on something, it pushes back.

From yahoo answers
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7 0

3 years ago

A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be

stepan [7]

Answer:

$p_2 = 664081 N/m^{2}$

Explanation:

from the ideal gas law we have

PV = mRT

$P = \rho RT$

$\rho = \frac{P}{RT}$

HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}

$\rho = \frac{7.00 10^{5}}{287(18+273)}$

$\rho = 8.38 kg/m^{3}$

We know by ideal gas law

$\rho = \frac{m_1}{V_1}$

$m_1 = \rho V_1 = 8.38 *2*10^{-3}$

$m_1 = 0.0167 kg$

for m_2

$m_2 = \rho V_i - V_removed$

$m_2 = 8.38*(.002 - 10^{-4})$

$m_2 = 0.0159 kg$

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

$\frac{p_2}{p_1}=\frac{m_2}{m_1}$

$p_2 = p_1 * \frac{m_2}{m_1}$

$p_2 =7*10^{5} * \frac {.0159}{0.0167}$

$p_2 = 664081 N/m^{2}$

8 0

3 years ago

4. As Juan is going to take a shower, the soap falls out of the soap dish on to the

LiRa [457]

The coefficient of friction between the soap and the floor is 0.081

If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.

We know that frictional force F = μN where μ = coefficient of friction between soap and floor.

So, μ = F/N

Since F = 40 N and N = W = 493 N,

μ = F/N

μ = 40 N/493 N

μ = 0.081

So, the coefficient of friction between the soap and the floor is 0.081

Learn more about coefficient of friction here:

brainly.com/question/13923375

5 0

3 years ago

A lamp draws a current of 0.1 A when it is connected to a 122-V source. (

Setler79 [48]

The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W

4 0

3 years ago

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,

maw [93]

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is

43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

5 0

3 years ago