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horrorfan [7]
3 years ago
11

A fisherman sees 7 wave crests go by in 10.0 s. The crests are 2.43 m apart. Find the period of the wave. (Unit = s)​

Physics
2 answers:
Gekata [30.6K]3 years ago
8 0

Velocity of the wave is 1.7 m/s

Explanation:

As we know that the frequency of the wave is defined as the number of waves crossing a fixed point per second

So here we will have

Also we know that the distance between two consecutive crest or two consecutive trough is known as wavelength

So here the wavelength of the wave is given as

now we have

nasty-shy [4]3 years ago
5 0

Answer:

hi

Explanation:

Wavelength=2.43

frequency=0.700 hz

t=1.43

v=1.70

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A mixture of two or more substance that can't be easily separated and it does make a new substance
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If the mixture cannot easily be separated then it would be called a compound.
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A satellite moves in a circular earth orbit that has a radius of 7.49 x 106 m. A model airplane is flying on a 24.1-m guideline
mina [271]

Answer:

Explanation:

Given

radius of satellite orbit r_1=7.49\times 10^6 m

And orbital velocity is given by

v=\sqrt{\frac{GM}{r}}

where M=mass of earth =5.98 \times 10^{24} kg

G=6.67\times 10^{-11}

v=\sqrt{\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}}{7.49\times 10^6}

v=7.29 \times 10^3 m/s

centripetal acceleration is given

a_c=\frac{v^2}{r}

a_c=\frac{(7.29\times 10^3)^2}{7.49\times 10^6}

a_c=7.095 m/s^2

For Model airplane

a_c=\frac{v^2}{r}

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4 0
4 years ago
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu
galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N

3 0
3 years ago
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Makovka662 [10]

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kd² = 2m (v² + 2g RL)

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d = √[2m (v² + 2g RL) / k]

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