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Mnenie [13.5K]
1 year ago
4

A block is hung vertically at the end of a spring. When the block is displaced and released, it moves in simple harmonic motion.

Which one of the following statements is true concerning the block
Physics
1 answer:
vazorg [7]1 year ago
6 0

The maximum acceleration of the block occurs when its velocity is zero.

<h3>What is harmonic motion?</h3>

Simple harmonic motion is a unique sort of periodic motion in mechanics and physics where the restoring force on the moving item is directly proportional to the amount of the object's displacement and acts in the direction of the object's equilibrium position. It is also referred to as SHM. If friction or any other energy dissipation is not there, it leads to an oscillation that lasts forever.

learn more about harmonic motion refer:

brainly.com/question/17315536

#SPJ4

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A net force applied to a 15.0 kg box produced an acceleration of 4.2 m/s2. If the same net force was applied to a 10 kg box, wha
igor_vitrenko [27]

Answer:

6.3\ m/s^2

Explanation:

Given that,

If the mass of a body is 15 kg and produced an 4.2 m/s², we need to find the acceleration if the mass is 10 kg and the same force is applied.

Force is given by :

F = ma

Since force is same

m_1a_1=m_2a_2\\\\a_2=\dfrac{m_1a_1}{m_2}\\\\a_2=\dfrac{15\times 4.2}{10}\\\\a=6.3\ m/s^2

So, if the mass is 10 kg, acceleration is 6.3\ m/s^2

8 0
3 years ago
a vector is 253 m long and points in a 55.8 degree direction, what’s the y and x- component of the vector?
BartSMP [9]

Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.

5 0
3 years ago
Help with both questions I’ll mark brainliest
creativ13 [48]

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5 0
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Which of these results in kinetic energy of an object? (1 point)
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A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns
Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

f'=\frac{V+V_{o}}{V-V_{s}} f  (1)

Where:

f=250,000 Hz is the actual frequency of the sound wave

f'=240,000 Hz is the "observed" frequency

V=340 m/s is the velocity of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which is the balloon

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}  (2)

V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}  (3)

V_{s}=-14.16 m/s (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time (t=4.8 s) given as data, we can find the distance:

d=V_{s}t (5)

d=(14.16 m/s)(4.8 s) (6)

Finally:

d=68 m (8) This is the distance of the balloon from the radar station

6 0
3 years ago
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