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RideAnS [48]
3 years ago
15

A flagpole consists of a flexible, 7.107.10 m tall fiberglass pole planted in concrete. The bottom end of the flagpole is fixed

in position, but the top end of the flagpole is free to move. What is the lowest frequency standing wave that can be formed on the flagpole if the wave propagation speed in the fiberglass is 27302730 m/s?
Physics
1 answer:
vladimir1956 [14]3 years ago
3 0

To develop this problem we require the concepts related to Frequency and their respective way of calculating it.

The formula to calculate the frequency is given by

f=\frac{V}{\lambda}

Where,

\lambda = wavelength

V= velocity

For fundamental mode, wavelength is equal to 4 time the length.

Then,

\lambda = 4L = 4*7.10m=28.4m

Replacing in the first equation,

f=\frac{2730}{28.4}\\

f= 96.12Hz

Therefore the frequency is 96.12Hz

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4 0
3 years ago
1. You serve a volleyball with a mass of 2.1kg. The ball leaves your
Darina [25.2K]

The kinetic energy is 945 joules.

Kinetic energy is the energy that an object has as a result of motion. It is defined as the effort required to accelerate a mass-determined body from rest to the indicated velocity.

The speed of an object or particle, which is a scalar quantity, is the size of the change in its location over time or the size of the change in its position per unit of time.

The mass of the volleyball is 2.1 kg.

The speed of the ball when the ball leaves the hand is 30 m/s.

m = 2.1 kg

v = 30 m/s

The kinetic energy of an object is given as:

KE = (1/2 ) × m × v²

KE = (1 / 2) × 2.1 kg × ( 30 m/s)²

KE = (1 / 2) × 2.1 kg × 30 m/s × 30 m/s

KE = 2.1 kg × 15 m/s × 30 m/s

KE = 945 J

Learn more about kinetic energy here:

brainly.com/question/8101588

#SPJ9

6 0
1 year ago
Referring to the graph below of data describing a cart pulled across a level surface. When a 4 N force pulls on the cart, the ca
LenKa [72]

Answer:gjv

Explanation:

4 0
2 years ago
Read 2 more answers
An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi
AVprozaik [17]

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

(1): The forces acting on the object must opposite each other. i.e. if force A acts at the right (or positive direction), force B will act at the left (or negative direction).

(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

7 0
2 years ago
Can someone solve this problem and explain to me how you got it​
evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

4 0
3 years ago
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