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3 years ago

Simple Harmonic Motion

Physics

1 answer:

Delvig [45]3 years ago

7 0

- The net force is greatest at the position of maximum displacement

- The net force is zero when at the equilibrium position

Explanation:

The motion of a spring is a Simple Harmonic Motion, in which the displacement of the end of the spring is given by a periodic function of the form

$x=Asin (\omega t)$

where A is the amplitude (the maximum displacement), and $\omega$ the angular frequency of the motion.

We can analyze the net force acting on the spring by looking at Hooke's law:

$F=kx$

where

F is the net force

k is the spring constant

x is the displacement

From the equation, we notice immediately that:

The net force is the greatest when the displacement x is the greates, so at the position in which the spring has maximum compression or stretching
The net force is zero when the displacement x is zero, so when the spring crosses the equilibrium position

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give an example of situation in which an automobile driver can have a centripetal acceleration but no tangential speed

Mars2501 [29]

There is no need for tangential acceleration when moving in a circle at a constant speed.

<h3>What is centripetal acceleration?</h3>

centripetal acceleration refers to the speed at which a body moves through a circle. Due to the fact that velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction), when a body travels in a circle, its direction is constantly changing, which causes a change in velocity, which results in an acceleration.

<h3>Which is an example of centripetal acceleration?</h3>

Centripetal acceleration occurs when you spin a ball on a string above your head. A car experiences centripetal acceleration when it is being driven in a circle. Additionally, a satellite in orbit around the Earth experiences centripetal acceleration.

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1 year ago

The acceleration due to gravity on the Moon's surface is

Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

$m = \frac{w}{g}$

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

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3 years ago

according to isaac newtons theory of gravitation the force exerted between two objects is dependent on- A- an objects weight. B-

puteri [66]

A an objects weight hope this helps! :D

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3 years ago

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and

lawyer [7]

1 kilometre=1000 metre

1 hour = 3600 second

$1\ km/hr=\frac{1000}{3600} m/s$

$1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

$35.0\ km/hr=35*\frac{5}{18} m/s$

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as $25\ km/hr^2$

$=\ 25*\frac{1000}{3600*3600} m/s^2$

$=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as $15\ km/hr^2$

$=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as $40\ km/hr^2$

$=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as- $60\ km/hr^2$

$=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

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A gun is fired parallel to the ground. at the same instant a bullet of equal size and mass next to the muzzle is released and dr

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Both hits the ground <u>at the same time</u> because they have <u>same vertical acceleration</u>

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<h3>What is vertical acceleration?</h3>

A vertical acceleration is typically one for which the direction of the vector is vertically upward, usually aligned with and opposite to the gravity vector. But this is a descriptive term, not a rigorous or technical term. A car may accelerate along a road and that would generally be assumed to be a horizontal.

The vector perpendicular to this direction, as perhaps a suspension motion over a bump, would be described as vertical even if it is not strictly vertical.

Note that acceleration is defined as the rate of change of the velocity vector. But the gravitation vector, ‘g’, generally vertically downward, is often denoted by what acceleration a mass in free fall (absent air resistance) would experience, i.e. the relationship between mass and weight.

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1 year ago