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Naily [24]
3 years ago
10

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

surements: a 2.50-kg stone thrown upwardfrom the ground at 15.0 {\rm m/s} returns to the ground in 9.00 {\rm s}; the circumference of Mongo at the equator is1.00×105 {\rm km} ; and there is no appreciable atmosphere onMongo.part AThe starship commander, Captain Confusion,asks for the following information: what is the mass ofMongo?m ={\rm kg}part bIf the Aimless Wanderer goes into acircular orbit 2.00×104 {\rm km} above the surface of Mongo, how many hours will ittake the ship to complete one orbit?t ={\rm h}
Physics
1 answer:
Setler [38]3 years ago
3 0

Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

         a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

          F = m a

         G m M / r² = m a

         M = a r² / G

Let's calculate

         M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

         F = m a

        G m M / R² = m a

Centripetal acceleration is

         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

Let's replace

        G m M / R² = m (2π R / T)² / R

        G M = R³ 4π² / T²

        T² = 4π² R³ / G M

       T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

       T² = 6.82 10¹⁶ / 3.33 10¹⁰

       T = √ (2,048 10⁶)

       T = 1.43 10³ s

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kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
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<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
3 years ago
ALL OF MY POINTS FOR THIS!
scoray [572]
Newton's 2nd law of motion:

                       Force  =  (mass) x (acceleration)

                                   =  (1,127 kg) x (6 m/s² forward)

                                   =  (1,127 x 6)  newtons forward

                                   =    6,762 newtons forward
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             Momentum  =  (mass) x (speed)

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3 0
3 years ago
A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle
Yuri [45]

Answer:

(a). Index of refraction are n_{red} = 1.344 & n_{violet} = 1.406

(b). The velocity of red light in the glass v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

The velocity of violet light in the glass v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

Explanation:

We know that

Law of reflection is

n_1 \sin\theta_{1} = n_2 \sin\theta_{2}

Here

\theta_1 = angle of incidence

\theta_2 = angle of refraction

(a). For red light

1 × \sin 56.6 = n_{red} × \sin 38.4

n_{red} = 1.344

For violet light

1 × \sin 56.6 = n_{violet} × \sin 36.4

n_{violet} = 1.406

(b). Index of refraction is given by

n = \frac{c}{v}

n_{red} = 1.344

v_{red} = \frac{c}{n_{red} }

v_{red} = \frac{3(10^{8} )}{1.344}

v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

v_{violet} = \frac{3(10^{8} )}{1.406}

v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

This is the velocity of violet light in the glass.

8 0
3 years ago
An example of an atom that has no change is one that has A. 1 proton, 2 electrons, and 3 neutrons B. 3 protons,2 electrons, and
Vikki [24]

Your question asks which is an example of an atom that has no change, or charge.

Your answer would be C). 2 protons, 2 electrons, and 1 neutron

These atoms would be known as a neutral atom.

In order for an atom to be neutral, it must have the same amount of protons and electrons.

We know that Protons are positive

We also know that Electrons are negative

In order for the atom to be neutral, they must have an equal amount of protons and neutrons to "cancel out"

In answer choice C, you would see that there are 2 protons and 2 electrons.

2- 0 = 0, in which allows this atom to be neutral.

Therefore, answer choice C. 2 protons, 2 electrons, and 1 neutron would be your answer.

4 0
3 years ago
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Two speakers, one directly behind the other, are each generating a 240-Hz sound wave. What is the smallest separation distance b
AveGali [126]

Answer:

The smallest separation distance between the speakers is 0.71 m.

Explanation:

Given that,

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz

Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m

To produce destructive interference at a listener standing in front of them,

d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m

So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.

3 0
3 years ago
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