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3 years ago

What is the period of vibration for a wave with a frequency of 100 hertz?

Physics

2 answers:

Taya2010 [7]3 years ago

5 0

Answer:

100 seconds

Hope this Helps :)

Dmitriy789 [7]3 years ago

3 0

The number of back-and-forth vibrations in a unit of time is an object's frequency. A complete back-and-forth vibration is one cycle. 1 Hz is one vibration or wave cycle per second!

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The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i

Stels [109]

Answer:

$T_1=0.24y$

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

$(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3$

Where $T_1$ and $T_2$ are the orbital periods of Mercury and Earth respectively. We have $D_1=0.39D_2$ and $T_2=1y$ . Replacing this and solving for

$T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y$

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3 years ago

The first step of the scientific method is to _______________.

chubhunter [2.5K]

The first step of the scientific method is to MAKE OBSERVATIONS

4 0

3 years ago

Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in

kolezko [41]

<h3>Answer;</h3>

A. 4

<h3>Explanation;</h3>

The period of a wave or periodic time is the time taken for a complete oscillation to occur. For example its is the time taken between two successive crests or troughs.
The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.
Frequency, measured in Hertz is given by the reciprocal of the periodic time.
Thus; Frequency or beats per second = 1/(1/4) = 4
Hence , 4 beats per second

6 0

3 years ago

What instrument is used to measure force?

katen-ka-za [31]

Barometer duhhhh what’s else a ruler

8 0

3 years ago

An electron moves with velocity v⃗ =(5.9i−6.4j)×104m/s in a magnetic field B⃗ =(−0.63i+0.65j)T. Determine the z-component of the

HACTEHA [7]

Answer:

Explanation:

Force on the electron = q ( v x B )

q = - 1.6 x 10⁻¹⁹

v = (5.9i−6.4j)×10⁴

B = (−0.63i+0.65j)

v x B = (5.9i−6.4j)×10⁴ x (−0.63i+0.65j)

= (3.835 - 4.032 ) x 10⁴ k

= - 1970 k

Force on the electron = q ( v x B )

= - 1.6 x 10⁻¹⁹ x -1970 k

= 3.152 x 10⁻¹⁶ k

z-component of the force on the electron

Fz = 3.152 x 10⁻¹⁶ N

7 0

3 years ago