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ycow [4]
3 years ago
5

A 2 kg, frictionless block is attached to a horizontal, ideal spring with spring constant 300 N/m. At t = 0 the spring is neithe

r stretched nor compressed and the block is moving in the negative direction at 12 m/s. (a) Find the amplitude of this oscillation.
Physics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

Explanation:

Given that,

Mass of block

M = 2kg

Spring constant k = 300N/m

Velocity v = 12m/s

At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0

xo = 0

It velocity is 12m/s at t=0

Then, it initial velocity is

Vo = 12m/s

Then, amplitude is given as

A = √[xo + (Vo²/ω²)]

Where

xo is the initial amplitude =0

Vo is the initial velocity =12m/s

ω is the angular frequency and it can be determine using

ω = √(k/m)

Where

k is spring constant = 300N/m

m is the mass of object = 2kg

Then,

ω = √300/2 = √150

ω = 12.25 rad/s²

Then,

A = √[xo + (Vo²/ω²)]

A = √[0 + (12²/12.5²)]

A = √[0 + 0.96]

A = √0.96

A = 0.98m

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KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement <em>s</em> = final position - initial position

s=ut+\frac{1}{2}at^{2}, where <em>u</em> is the initial velocity, <em>a</em> is the

constant acceleration and <em>t</em> is the time

So we can find the final velocity by using the rule:

final position - initial position = ut+\frac{1}{2}at^{2}

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = (-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

<em>That means the car is at 8.1625 meters in opposite direction</em>

<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>

4 0
3 years ago
1.Predict the frequency of a tuning fork that emits a sound with a wavelength of 0.385 m.
lina2011 [118]

Answer:

1.) Frequency F = 890.9 Hz

2.) Wavelength (λ) = 0.893 m

Explanation:

1.) Given that the wavelength = 0.385m

The speed of sound = 343 m / s

To predict the frequency, let us use the formula V = F λ

Where (λ) = wavelength = 0.385m

343 = F × 0.385

F = 343/0.385

F = 890.9 Hz

2.) Given that the frequency = 384Hz

Using the formula again

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λ = V/F

Wavelength (λ) = 343/384

Wavelength (λ) = 0.893 m

The two questions can be solved with the use of formula

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zavuch27 [327]
The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.

Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.

Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
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What happens when calcium reacts with chlorine?
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Answer:

Calcium chloride in an ionic compound. The calcium atom loses two electrons, and each chlorine atom gains one electron, to form a full outer shell of electrons.

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Answer:

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