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salantis [7]
4 years ago
15

A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con

stant for B is one quarter of the spring constant for A.B.The spring constant for B is 4 times the spring constant for A.C.The spring constant for B is half of the spring constant for A.D.The spring constant for B is 1.41 times the spring constant for A.E.The spring constant for B is twice the spring constant for A.
Physics
1 answer:
Nataliya [291]4 years ago
3 0

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

T_B = 2T_A

As T = 2\pi\sqrt{\frac{m}{k}}, and the 2 springs have the same mass

2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}

\sqrt{k_A} = 2\sqrt{B}

k_A = 4k_B

k_B = k_A/4

So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.

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She should explain that the Sun is made up of gaseous layers that surround an iron core.

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3 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 14 s until its motor stops. Disreg
telo118 [61]

Answer:

The maximum height of the rocket will be 1.0 × 10⁴ m.

Explanation:

Hi there!

The height of the rocket at time "t" can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²  (when the rocket has an upward acceleration)

y = y0 + v0 · t + 1/2 · g · t²  (after the motor of the rocket stops)

Where:

y = height.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to the motors.

g = acceleration due to gravity.

The velocity of the rocket can be calculated as follows:

v = v0 + a · t  (while the motor is running)

v = v0 + g · t  (after the motor stops)

Where "v" is the velocity of the rocket at time "t".

The rocket rises with upward acceleration for 14 s. After that, the rocket starts being accelerated in the downward direction due to gravity. But it will continue going up after the motor stops because the rocket has initially an upward velocity that will be reduced until it becomes zero and the rocket starts to fall.

Let´s find the height reached by the rocket while it was accelerated in the upward direction (the origin of the frame of reference is located at the launching point and upward is the positive direction):

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · 14 s + 1/2 · 28 m/s² · (14 s)²

y = 2.7 × 10³ m

Now let´s find the velocity reached in that time:

v = v0 + a · t

v = 28 m/s² ·14 s

v = 3.9 × 10² m/s

Now, let´s find the maximum height reached by the rocket using the equations of height and velocity after the motor stops:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Notice that now v0 and y0 will be the velocity and height reached while the rocket was being accelerated in the upward direction, respectively.

Let´s find at which time the rocket reaches its maximum height. With that time, we can calculate the max-height.

At the maximum height, the velocity of the rocket is zero, then:

v = v0 + g · t

0 = 3.9 × 10² m/s - 9.8 m/s² · t

-3.9 × 10² m/s/ -9.8 m/s² = t

t = 40 s

After the motor stops, it takes the rocket 40 s s to reach the maximum height.

Using the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 2.7 × 10³ m +  3.9 × 10² m/s · 40 s - 1/2 · 9.8 m/s² · (40 s)²

y = 1.0 × 10⁴ m

The maximum height of the rocket will be 1.0 × 10⁴ m

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Identify the following as
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