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2 years ago

M A sinusoidal wave on a string is described by the wave function

Physics

1 answer:

IceJOKER [234]2 years ago

5 0

The frequency of the wave is determined as 7.96 Hz.

<h3>Frequency of the wave</h3>

The frequency of the wave is calculated as follows;

y = A sin(ωt - kx)

where;

A is amplitude of the wave
ω is angular speed of the wave

ω = 2πf

f = ω/2π

f = (50)/(2π)

f = 7.96 Hz

Thus, the frequency of the wave is determined as 7.96 Hz.

Learn more about frequency of waves here: brainly.com/question/6297363

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The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of

PolarNik [594]

The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions

7 0

3 years ago

Read 2 more answers

A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its ra

mr_godi [17]

<h2>Answer:</h2>

The rate of deceleration is -0.14 $m/s^{2}$

<h2>Explanation:</h2>

Using one of the equations of motion;

v = u + at

where;

v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)

u = initial velocity of the boat = 25m/s

a = acceleration of the boat

t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes

Convert the time t = 3 minutes to seconds;

=> 3 minutes = 3 x 60 seconds = 180seconds.

Substitute the values of v, u, t into the equation above. We have;

v = u + at

=> 0 = 25 + a(180)

=> 0 = 25 + 180a

Make a the subject of the formula;

=> 180a = 0 - 25

=> 180a = -25

=> a = -25/180

=> a = -0.14 $m/s^{2}$

The negative value of a shows that the boat is decelerating.

Therefore, the rate of deceleration of the speed boat is 0.14 $m/s^{2}$

5 0

3 years ago

A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

6 0

3 years ago

Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows

Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle $\theta _c = sin^{-1} (\frac{n}{n_{slab}} )$

$sin \theta _ c =\frac{n}{n_{slab}}$

According to Snell's law of refraction:

$n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)$

At point P ; $90 - \theta _2 \leq \theta _c$

$\theta _2 = 90 - \theta _c$

Therefore:

$n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}$

Then maximum value of refractive index n of the fluid is:

$n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }$

$n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }$

n = 1.4266

3 0

3 years ago

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma

mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

${}$ ⇩

[m = 20 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

$I_i \cdot \omega _i = I_f \cdot \omega _f$

The moment of inertia of the merry-go-round, $I_m$ = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²· $\omega _i$ = (0.5·M·R² + m·r²)· $\omega _f$

Given that 0.5·M·R²· $\omega _i$ is constant, as the value of m·r² increases, the value of $\omega _f$ decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

Learn more here:

brainly.com/question/15188750

6 0

2 years ago