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solmaris [256]
3 years ago
10

A ball is tossed straight up from the surface of a small, sphericalasteroid with no atmosphere. The aball rises to a height equa

l tothe asteroid's raidus and then falls straight down toward thesurface of the asteroid.
1. What forces act on the ball while it is on the way up?

1) only a decreasing gravitational force that acts downward.
2) only an increasing gravitational forces that acts downward
3) only a constant gravitation force that acts downward
4) both a constant gravitational force that acts downward and adecreasing force that acts upward
5) no forces act on the ball

2. the acceleration of the ball at the top of its path is1) at its max vlue for the ball's flightb) Equal to the accleration at the surface of the asteroidc) equal to .5 the acceleration at the surface of the asteroidd) equal to 1/4 the acceleration at the surface of theasteroide) zero
Physics
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

1. 3) only a constant gravitation force that acts downward

2. b) Equal to the accleration at the surface of the asteroidc)

Explanation:

1.

  • In absence of atmosphere, the force that will act on the steroid is the force of gravity due to asteroid.
  • Gravitational force is a long range force acting always attractive in nature.
  • It is a contact-less field force which does not requires a medium.

Mathematically given as:

F=G.\frac{M_1.M_2}{R^2}

where:

G = gravitational constant

M_1\ \&\ M_2 are the mass of the two objects

R= radial distance between the two objects.

2.

The acceleration of the ball at the top height of the path is still under the influence of the gravity of the two masses so it will be equal to the acceleration due to gravity at the surface of the asteroid. Acceleration always remains constant.

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Answer: D. it has been demonstrated to be without exception under certain stated conditions.

Explanation:

A <u>Law</u> is an affirmation (something established) based on repeated long-term observation of a phenomenon that has been studied and verified.  

That is: A law is present in all known theories and therefore is considered universal. In addition, a law can not be refuted, nor changed, because its precepts have been proven through various studies.  

<u>Unlike theory</u>, which is the set of rules and principles that describe and explain a particular phenomenon and <u>is subject to changes as new evidence emerges that gives meaning to it.  </u>

Then, based on what is explained above, the law of universal gravitation is a statement that exists because it was rigorously tested and verified, therefore it can not be refuted.

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Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
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Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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