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maxonik [38]
1 year ago
6

A 297 g block is connected to a light spring with spring constant 4.34 N/m, and displaced 7.45 cm from equilibrium. It is then r

eleased and allowed to oscillate in simple harmonic motion. What is the maximum acceleration of the block
Physics
1 answer:
vagabundo [1.1K]1 year ago
4 0

Answer:

x = A sin ω t       describes the displacement of the particle

v = A ω cos ω t

a = -A ω^2 sin ω t        

a (max) = -A ω^2     is the max acceleration (- can be ignored here)

ω = (K/ m)^1/2        for SHM

F = - K x^2     restoring force of spring

K = 4.34 / .0745^2 = 782 N / m

ω = (782 / .297)^1/2 = 51.3 / sec

a (max) = .0745 * 782 / .297 = 196 m / s^2

You might be interested in
If a ball has a velocity of 18 m/s, how far does it travel in 15 seconds
Stolb23 [73]

Answer:

The answer is

<h2>270 m</h2>

Explanation:

To find the distance when given the velocity and time we use the formula

<h3>distance = velocity × time</h3>

From the question

velocity of the ball = 18 m/s

time = 15 s

So the distance is

distance = 18 × 15

We have the final answer as

<h3>270 m</h3>

Hope this helps you

7 0
3 years ago
Statistical time division multiplexing does not require the capacity of the circuit to be equal to the sum of the combined circu
aleksklad [387]

Answer:

The answer is True

Explanation:

Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.

In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.

Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.

7 0
3 years ago
Calculate Vector component in Y if the hypotenuse is 32 and angle is 45
Lerok [7]

Answer:

The correct option is;

c. 22.6

Explanation:

The given parameters are;

The hypotenuse of the vector = 32

The angle of the vector = 45°

Therefore, the vector component in the y-axis is given as follows;

v_y = v \times sin(\theta)

Substituting the values from the question gives;

v_y = 32 \times sin(45^{\circ}) \approx 22.6

The vector component in the y-axis, v_y, is approximately 22.6.

8 0
2 years ago
Physics question, please help?
Ludmilka [50]

0.4823 m/s

The initial velocity u1 of the ball=0

From the law of conservation of linear momentum.

m1u1+m2u2=m1v1+m2v2

(160×0)+(170×u1)=(160×0.3)+(170×0.2)

u1=0.4823m/s

6 0
3 years ago
A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe
joja [24]

Answer:

The Doppler Effect is given by the following relation;

f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

v_o = The velocity of the observer

v_s = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(ii) When the source is receding from the observer, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

(1) Given that (v + v_s) > v, therefore, v < (v + v_s), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(1) Given that (v - v_s) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0
3 years ago
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