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3 years ago

A plane progressive

Physics

1 answer:

kiruha [24]3 years ago

4 0

Answer:

Amplitude, A = 5 m

Explanation:

Let a progressive wave is given by equation :

$y=5\sin (100\pi t-0.4\pi x)$ .....(1)

The general equation of a progressive wave is given by :

$y=A\sin (\omega t-kx)$ ....(2)

Here,

A is the amplitude of the wave

$\omega$ is the angular frequency

k is propagation constant

We need to find the amplitude of the wave.

If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.

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Two charges repel one another at 5N, if the distance between them is 2m and the force increases to 25N, what is the new distance

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Answer:

Explanation:

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3 years ago

How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would

professor190 [17]

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

$Circumference=2\pi R$

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$v=\frac{Distance}{Time}$

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

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5 0

4 years ago

Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing

Juliette [100K]

Answer: MR²

is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

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3 years ago

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter

expeople1 [14]

Answer: $56.87m/s^{2}$

Explanation:

If we make an analysis of the net force $F_{net}$ of the rock that was thrown upwards, we will have the following:

$F_{net}=F_{up}-W$ (1)

Where:

$F_{up}=200N$ is the force with which the rock was thrown

$W$ is the weight of the rock

Being the weight the relation between the mass $m=3kg$ of the rock and the acceleration due gravity $g=9.79m/s^{2}$ :

$W=m.g=(3kg)(9.79m/s^{2})$ (2)

$W=29.37 N$ (3)

Substituting (3) in (1):

$F_{net}=200N-29.37 N$ (4)

$F_{net}=170.63 N$ (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

$F_{net}=m.a$ (6)

Finding the acceleration $a$ :

$a=\frac{F_{net}}{m}$ (7)

$a=\frac{170.63 N}{3kg}$ (8)

Finally:

$a=56.87m/s^{2}$

3 0

3 years ago

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto

Fed [463]

Answer:

a) a = 34.375 m / s², b) v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

v_f = $\frac{x-x_1}{t}$

we substitute the values

v_f = $\frac{ 6600 -x_1}{4}$

The initial part of the movement is carried out with acceleration

v_f = v₀ + a t

x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

x₁ = ½ a t²

v_f = a t

we substitute the values

x₁ = 1/2 a 16²

x₁ = 128 a

v_f = 16 a

let's write our system of equations

v_f = $\frac{6600 - x_1}{4}$

x₁ = 128 a

v_f = 16 a

we substitute in the first equation

16 a = $\frac{6600 -128 a}{4}$

16 4 a = 6600 - 128 a

a (64 + 128) = 6600

a = 6600/192

a = 34.375 m / s²

b) let's find the time to reach this height

x = ½ to t²

t² = 2y / a

t² = 2 5100 / 34.375

t² = 296.72

t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

v_f = 16 a

v_f = 16 34.375

v_f = 550 m / s

8 0

3 years ago