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marusya05 [52]
1 year ago
6

A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a

nd period t. A student grabs the mass and brings it to rest before starting experiment 2. In experiment 2, the mass is set to oscillate with a larger amplitude of 3a. What is the period of the oscillation in experiment 2?.
Physics
1 answer:
Trava [24]1 year ago
6 0

Time period remains the same in both the experiment as change in amplitude does not affect time period.

What are the factors on which time period depends in SHM?

Time period is given by:

T=2\pi \sqrt{\frac{m}{k} }

where,

T = time period

m = mass

k = spring constant

In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.

What is Spring Constant?

A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.

Learn more about SHM here:

brainly.com/question/20885248

#SPJ4

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What is the passengers velocity on the vertical drop of the tower of terror at Disney world if the distance for the drop is 50m,
andrey2020 [161]

Assuming constant acceleration due to gravity of a=9.8\,\dfrac{\mathrm m}{\mathrm s^2}, and assuming the passengers start at rest, so that v_0=0\,\dfrac{\mathrm m}{\mathrm s}, we have

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7 0
3 years ago
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to t
WINSTONCH [101]

Complete question is;

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer:

B_net = 50 × 10^(-7) T

Explanation:

We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).

This means that the second wire is 4 m in length on the positive y-axis.

Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.

This means that the position we want to find is half the length of the second wire.

Thus, at this point the net magnetic field is given by;

B_net = √[(B1)² + (B2)²]

Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.

Now, formula for magnetic field due to very long wire is;

B = (μ_o•I)/(2πR)

Thus;

B1 = (μ_o•I_1)/(2πR_1)

Also, B2 = (μ_o•I_2)/(2πR_2)

Now, putting the equation of B1 and B2 into the B_net equation, we have;

B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]

Now, factorizing out some common terms, we have;

B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]

Now,

μ_o is a constant and has a value of 4π × 10^(−7) H/m

I_1 = 30 A

I_2 = 40 A

Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.

Thus;

R_1 = 2 m

R_2 = 2 m

So, let's calculate B_net.

B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]

B_net = 50 × 10^(-7) T

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3 years ago
Light from a 560 nm monochromatic source is incident upon the surface of fused quartz (n = 1.56) at an angle of 60°. What is the
MaRussiya [10]

Answer:

The angle of reflection is "60°".

Explanation:

The given values are:

Light from monochromatic source,

= 560 nm

Angle of incidence,

= 60°

Surface of fused quartz (n),

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Whenever a light ray was indeed occurring at a flat surface, it should be the law or concept of reflection which contains this same rays of light, the reflected ray as well as the "normal" ray at either the mirror surface.

According to the above law,

⇒  Angle \ of \ incidence=Angle \ of \ reflection

then,

⇒  Angle \ of \ reflection=60^{\circ}

3 0
3 years ago
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