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lbvjy [14]
3 years ago
8

If an ocean wave passes a stationary point every 5 s and has a velocity of 10 m/s, what is the wavelength of the wave

Physics
1 answer:
Rainbow [258]3 years ago
8 0

Answer:

As Per Provided Information

Velocity of wave v is 10m/s

These ocean wave passes a stationary point every 5 s ( It's time period)

First we calculate the frequency of ocean wave .

<u>Using</u><u> Formulae</u>

\blue{\boxed{\bf \:  \nu =  \cfrac{1}{v}}}

here

v is the velocity of wave .

\sf\longrightarrow \nu \:  =  \cfrac{1}{10}  \\  \\ \\  \sf\longrightarrow \nu \:  = 0.1Hz

Now calculating the wavelength of the wave .

<u>Using </u><u>Formulae </u>

\boxed{ \bf \lambda =  \cfrac{v}{ \nu}}

Substituting the value and we obtain

\sf \longrightarrow \lambda \:  =  \cfrac{10}{0.1}  \\  \\  \\ \sf \longrightarrow \lambda \:  =   \cancel\cfrac{10}{0.1}  \\  \\  \\ \sf \longrightarrow \lambda \:  =100m

<u>Therefore</u><u>,</u>

  • <u>Wavelength </u><u>of </u><u>the </u><u>wave </u><u>is </u><u>100 </u><u>metres</u><u>.</u>
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Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

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Answer:

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Answer:

Explanation:

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The apparent weight of the pilot would be 619.48 N and the speed he would experience at weightlessness is 192. 97 m/s

<h3>What is apparent weight?</h3>

Apparent weight is a property of object that co-relates to the heaviness of the object.

It differs from the weight of an object as a result of the in-balance between the force acting on the object and an equal but opposite force

<h3>How to calculate the apparent weight</h3>

Formula:

Apparent weight = Real weight (mg) - m v∧2 ÷ r

Where m = mass

g = acceleration due to gravity = 9.8 m/s²

v = constant velocity = 2. 10 × 10∧2 m/s

r = radius = 3.80 × 10∧3 m

Note that weight = mg

m = weight ÷ g = 623 ÷ 9.8 = 63. 57 kg

a. Apparent wieight = 623 - (63. 57 × 2.10 × 10∧2 × 2.10 × 10∧2) ÷ 3.80 × 10∧3)

= 623 - (63.57 × 210) ÷ 3800

= 623 - 13377 ÷ 3800

= 623 - 3. 52

= 619.48 N

b. At weightless, weight = 0

Speed, V = √g× r = √ 9.8 × 3800 = √37,240 = 192. 97 m/s

Hence, The apparent weight of the pilot would be 619.48 N and the speed he would experience at weightlessness is 192. 97 m/s

Read more on apparent weight here:

brainly.com/question/11180653

#SPJ1

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