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timofeeve [1]
3 years ago
13

A ball on a string is swung around in a horizontal circle. The centripetal force

Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

0.60 N, towards the centre of the circle

Explanation:

The tension in the string acts as centripetal force to keep the ball in uniform circular motion. So we can write:

(1)

where

T is the tension

m = 0.015 kg is the mass of the ball

is the angular speed

r = 0.50 m is the radius of the circle

We know that the period of the ball is T = 0.70 s, so we can find the angular speed:

And by substituting into (1), we find the tension in the string:

And in an uniform circular motion, the centripetal force always points towards the centre of the circle, so in this case the tension points towards the centre of the circle.

Explanation:

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Zigmanuir [339]

Answer: The work is 1863 N*m

Explanation:

We can define work as:

W = F*d

Where F is the force that the mover needs to apply to the refrigerator, and d is the distance that the refrigerator is moved.

To move the refrigerator, the minimal force that the mover needs to do is exactly the friction force (In this case, the refrigerator will move with constant speed).

Then we will have:

F = 230 N

and the distance is 8.1 meters, then the work will be:

W = 230N*8.1 m = 1863 N*m

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A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m + t/(0.20 s))], where x is in m and t in s.
Len [333]

Corrected and Formatted Question:

A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.

(a) In what direction is this wave traveling?

(b) What are the wave speed, frequency, and wavelength?

(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?

Answer:

The wave is travelling in the negative x direction

The wave speed = 12.0m/s

The frequency = 5Hz

The wavelength = 2.4m

The displacement at t = 0.50s and x = 0.20m is -0.029m

Explanation:

The general wave equation is given by;

y(x, t) = y cos (2\pi(x/λ) - 2\pift)    --------------------------------(i)

Where;

y(x, t) is the displacement of the wave at position x and a given time t

y = amplitude of the wave

f = frequency of the wave

λ = wavelength of the wave

Given;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]   ------------------(ii)

Which can be re-written as;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))]  -------------(iii)

Comparing equations (i) and (iii) we have that;

=> 2π(x/(2.4 m) = 2π(x/λ)

=> λ = 2.4m

Therefore the wavelength of the wave is 2.4m

Also, still comparing the two equations;

=> 2π(t/(0.20 s) = 2πft

=> f = 1 / 0.20

=> f = 5Hz

Therefore the frequency of the wave is 5Hz

To get the wave speed (v), it is given by;

v = f x λ

Where f = 5Hz and λ = 2.4m

=> v = 5 x 2.4

=> v = 12.0m/s

Therefore, the speed of the wave is 12.0m/s

At t = 0.50s and x = 0.20m;

The displacement, y(x,t) of the string wave is given by

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Convert the amplitude of 3.0cm to m</em>

=> 3.0cm = 0.03m

<em>Substitute this back into the equation</em>

=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Substitute the values of t and x into the equation above;</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]

<em>Carefully solve the equation</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]

=> y(x, t) = (0.03m) × cos[0.08π + 5π]

=> y(x, t) = (0.03m) × cos[5.08π]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × -0.9684

=> y(x, t) = 0.029m

Therefore the displacement at those points is -0.029m

Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.

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What is the relation between the direction of electric field and equipotential line at the same point
gavmur [86]

Answer:

The direction of electric field and equipotential line at the same point are always PERPENDICULAR TO THE ELECTRIC FIELD.

Explanation:

Equipotential surface is a three dimensional part of equipotential lines.

Equipotential lines are a type of contour lines that is use to trace lines that have the same altitude on the map and the altitude is the electric potential.

Equipotential lines are always perpendicular to electric potential because the lines creates three dimension equipotential surface.

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