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Alexeev081 [22]
3 years ago
13

Calculate the position of the center of mass of the following pairs of objects. Use a coordinate system where 0 is at the center

of the more massive object. Give your answer not in meters but as a fraction of the radius as requested. You can find the data you need in the inside of the back cover of your textbook. (a) The Earth and the Moon. Give the answer as a fraction of the earth's radius.
Physics
1 answer:
vlada-n [284]3 years ago
8 0

Answer:

the center of mass is  Rcm = 0.725 R Earth

Explanation:

The concept of center of mass is of great importance in all physics, since on this all external forces of the system can be considered applied, in the collision analysis it gives an ideal reference system for resolution

The equation for the center of mass is

        Rcm = ∑ mi ri / Mt

With Rcm and ri being vector quantities, "m" is body mass and MT total mass

If we write this equation for our system we have

       Rcm = (M1 r1 + M2 r2) / (M1 + M2)

M1 and M2 are the mass of the Earth and the Moon respectively, r1 and r2 are the distances measured from our reference system. As they indicate that we place the most massive reference system we place it on planet Earth.  The data taken a textbook.

       

        M1 = 5.98 10²⁴  Kg

        M2 = 7.36 10²² Kg

       Distance from Earth to the moon on average 3.84 10⁸ m

How the coordinate system is placed on Earth

 

        r1 = 0

        r2 = 3.84 10 8 m

      Rcm = 0 + 7.36 10²²  3.84 10⁸ / (5.98 10²⁴ + 7.36 10²²)

R cm = 2.826 10³¹ / (598 10²² + 7.36 10²²)

Rcm = 2.826 10³¹ / 611.34 10²²

R cm = 4.62 10⁶  m

we are asked to give this distance as a fraction of the radius of the Earth, we divide the two quantities

      R earth = 6.37 10⁶ m

     Fraction = R cm / R Earth

     Fraction = 4.62 10⁶ / 6.37 10⁶

     Fraction = 0.725

In summary the center of mass is

     Rcm = 0.725 R Earth

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Answer:

Gamma beta alpha

Explanation:

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Solve for the length of the inclined plane if the angle equals 19.45 degrees.
mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

The length of the inclined plane is the hypotenuse of the triangle. The length of the inclined plane can be found using trigonometric ratios.

height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

h = 4 / 0.33298412235

h = 12.0125847796

h = 12 ft

Therefore, the length of the inclined plane is approximately 12 ft

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Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

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3 years ago
A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
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a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the package

u is the initial velocity

t is the time

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We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

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Natali5045456 [20]

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

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q = -5nC - 15nC

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Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

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