Ask question

Ask question

All categories

3 years ago

12

Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha

t is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.

1 answer:

ikadub [295]3 years ago

8 0

We have that the spring constant is mathematically given as

$k=2.37*10^{11}N/m$

Generally, the equation for angular velocity is mathematically given by

$\omega=\sqrt{k}{m}$

Where

k=spring constant

And

$\omega =\frac{2\pi}{T}$

Therefore

$\frac{2\pi}{T}=\sqrt{k}{n}$

Hence giving spring constant k

$k=m((\frac{2 \pi}{T})^2$

Generally

Mass of earth $m=5.97*10^{24}$

Period for on complete resolution of Earth around the Sun

$T=365 days$

$T=365*24*3600$

Therefore

$k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2$

$k=2.37*10^{11}N/m$

In conclusion

The effective spring constant of this simple harmonic motion is

$k=2.37*10^{11}N/m$

For more information on this visit

brainly.com/question/14159361

You might be interested in

Need some help please answer please

yuradex [85]

Answer:

Its not d or e

Explanation:

8 0

2 years ago

Select all of the following that describes momentum, p [mark all correct answers]

Inga [223]

Answer:

c. Momentum is the product of mass and velocity

e. Momentum is a vector quantity

g. Momentum has unit of kgm/s

Explanation:

Linear momentum P

P = m .v

m =mass

v=Velocity

If mass take in kg and velocity is in m/s then momentum p will be in kg.m/s.

1. momentum is the product of velocity and mass.

2.Momentum is a vector quantity.

3.Momentum have kg.m/s unit.

So the following option are correct.

c. Momentum is the product of mass and velocity

e. Momentum is a vector quantity

g. Momentum has unit of kgm/s.

Note-

1.Joule is the unit of energy.

2.One-half the product of mass and the square of the object's speed is known as kinetic energy.

7 0

3 years ago

In this photograph, a soccer player is about to kick the ball. Use the situation to explain that when two objects interact, the

Makovka662 [10]

hey you look nice (pic).

According to Newton’s first law, if no force is applied to a ball, it will continue moving at the same speed and direction as it did before. When we put the ball on the grass it stays in its place, namely it stays in zero motion since no force is applied to it. However, after we kick the ball, it will continue moving in the direction we kicked it. Its speed will drop gradually, due to friction (a force applied on the ball in the opposite direction to its motion), but the direction of its motion will remain the same.

According to Newton’s second law, a force applied to an object changes that object’s acceleration – namely, the rate at which the speed of the object changes. When we kick the ball, the force we apply to it causes it to accelerate from a speed of 0 to a speed of dozens of kilometers per hour. When the ball is released from the foot, it begins to decelerate (negative acceleration) due to the force of friction that is exerted upon it (as we observed in the previous example). If we were to kick a ball in outer space, where there is no friction, it would accelerate during the kick, and then continue moving at a constant speed in the direction that we kicked at, until it hits some other object or another force is applied to it.

8 0

3 years ago

Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object

ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.

Can you solve this system of equations seeing them like this, or do you need more help?

6 0

3 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago