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dangina [55]
3 years ago
5

A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its

speed is _______ and its acceleration is _______.0,0max,max0,maxmax,o
Physics
2 answers:
Leni [432]3 years ago
5 0

Answer:

Maximum; 0

Explanation:

As the mass moves through the equilibrium position, it is moving at its fastest speed; and there's no restoring force at that moment, so the acceleration is zero.

hram777 [196]3 years ago
3 0

Answer:

The speed is maximum and the acceleration is zero

Explanation:

- The speed of the mass in simple harmonic motion can be found by using the law of conservation of energy. In fact, the total mechanical energy of the mass-spring system is sum of kinetic energy and elastic potential energy:

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

where

m is the mass

v is the speed

k is the spring constant

x is the displacement

As we can see, when the displacement is zero (x=0), the term representing the kinetic energy is maximum, so v (the speed) is also maximum.

- The acceleration of the mass in simple harmonic motion is proportional to the restoring force acting on the mass, which is given by Hook's law

a \propto F = -kx

where

k is the spring constant

x is the displacement

When x = 0, F = 0, so the net force acting on the mass is zero. Therefore, this also means that the acceleration of the mass is also zero: a = 0.

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Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
A person walks 9 meters to the right and 8 meters to the left what's the distance and displacement​
nordsb [41]
The distance is 17 and the displacement is 1
5 0
2 years ago
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The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c
Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0
3 years ago
How long will the ball be in the air if the cliff is 120 m tall and the ball falls to the base of the cliff?
Readme [11.4K]

Answer:

Explanation:

4.95s≈5s

Use equation :

h=G*t²/2

h=120m ----hight of cliff

G=9.81m/s²

t=?

-----------------------

h=G*t²/2

120m=9.81m/s²*t²/2

240m=9.81m/s²*t²

t²=240m/ 9.81m/s²

t²=24.46s

t=√24.46s²

t=4.95 s≈5s

3 0
4 years ago
Without Newton’s third law of motion, what would an object sitting on a table do?
Angelina_Jolie [31]

Newton's third law: If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself.

8 0
3 years ago
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