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katen-ka-za [31]
1 year ago
15

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m

/s . At t=0, an element of the string at x = 0 has a transverse position of 2.00 cm and is traveling downward with a speed of 2.00 m/s.(a) What is the amplitude of the wave?
Physics
1 answer:
Stolb23 [73]1 year ago
3 0

The amplitude of the wave is determined as 0.0215 m.

<h3>What is amplitude of a wave?</h3>

The amplitude of a wave is the maximum displacement of the wave.

The amplitude of the wave is calculated as follows;

<h3>Amplitude of the wave</h3>

A cos(Φ) = 0.02 ---- (1)

-80π A sin(Φ) = - 2

A sin(Φ) = 1/40π  ----- (2)

Divide (2) by (1)

tan(Φ) = 1/(0.02 x 40π)

tan(Φ) = 1/(0.8π)

Ф = tan⁻¹( 1/(0.8π))

Ф = 0.3787

A cos (0.3787) = 0.02

A (0.929) = 0.02

A = 0.0215 m

Thus, the amplitude of the wave is determined as 0.0215 m

Learn more about amplitude of wave here: brainly.com/question/19036728

#SPJ4

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sergiy2304 [10]

Answer:

10 cm

Explanation:

water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B protection is 10 cm . the bar graph attached in the shows that the lengths of yellow and brown bars differ the most at 10 cm.

Hence,The biggest difference between the survival rates of UV-B protected and unprotected can be seen at the depth of 10 cm.

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3 years ago
5. An object accelerates from 10km/h at a rate of 5m/s2. What distance has the object traveled when it
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6 0
3 years ago
a block accelerates at 3.3 m/s^2 down a plane inclined at an angle of 27°. find the kinetic friction coefficient between the blo
finlep [7]

Answer:

0.13

Explanation:

Draw a free body diagram.  There are three forces:

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Normal force N pushing perpendicular to the incline.

Friction force Nμ pushing parallel up the incline.

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel down the incline:

∑F = ma

mg sin θ − Nμ = ma

mg sin θ − mgμ cos θ = ma

g sin θ − gμ cos θ = a

gμ cos θ = g sin θ − a

μ = (g sin θ − a) / (g cos θ)

Given a = 3.3 m/s² and θ = 27°:

μ = (9.8 m/s² sin 27° − 3.3 m/s²) / (9.8 m/s² cos 27°)

μ = 0.13

3 0
3 years ago
A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The
mezya [45]

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

v = v_{o}+a\cdot t

Where:

v_{o} - Initial velocity, measured in meters per second.

v - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t - Time, measured in seconds.

Now, we clear acceleration within expression:

a = \frac{v-v_{o}}{t}

Initial and final velocities are now converted from kilometers per hour into meters per second:

v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v_{o} = 30\,\frac{m}{s}

v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v = 37.5\,\frac{m}{s}

If we know that t = 3.5\,s, then, the average acceleration of the car is:

a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}

a = 2.143\,\frac{m}{s^{2}}

The average acceleration of the car is 2.143 meters per square second.

8 0
3 years ago
A 1500 kg car moving with a speed of 20 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of
storchak [24]

Answer:

-100000 N.

Explanation:

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Where F = Average force exerted on the car, m = mass of the car, acceleration of the car, a = acceleration of the car.

a = (v-u)/t..................... Equation 2

Where v = Final velocity, u = Initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............. Equation 3

Given: m = 1500 kg, u = 20 m/s, v = 0 m/s (brought to rest), t = 0.3 s.

Substitute into equation 3

F = 1500(0-20)/0.3

F = 1500(-20)/0.3

F = -100000 N.

Note: The negative sign is due to the fact that the force exerted on the car by the pole is equal and opposite the force of the car.

7 0
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