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S_A_V [24]
3 years ago
11

A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?

Physics
2 answers:
Veseljchak [2.6K]3 years ago
8 0

idk it's wrong chose ...remains constant...

Lady_Fox [76]3 years ago
7 0
Vertical component of velocity at the highest point becomes zero but horizontal component of velocity remains constant throughout the projectile motion.
Hope this helps!
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Why is gravity important in the solar system
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A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
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To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

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3 years ago
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le
expeople1 [14]

Answer:

The  value  is   m \approx   310

Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

    The  focal length of the eyepiece is  f_e  =  2.0 \  cm

    The  tube length is  L  =  25 \  cm

Generally the magnitude of the overall magnification is mathematically represented as

            m =  m_o  *  m_e

Where  m_o is the objective magnification which is mathematically represented as

        m_o  =  \frac{L}{f_o }

=>      m_o  =  \frac{25}{1 }

=>      m_o  =  25

m_e is the eyepiece magnification which is mathematically evaluated as

     m_e  =  \frac{L }{f_e }

     m_e  =  \frac{25 }{ 2}

      m_e  =  12.5 \  cm

So

    m =  25 * 12.5

     m \approx   310

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3 years ago
What is the action force and reaction force of two bumper cars collide
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<span>action is the one car hitting the other, reaction is the other car being pushed away</span>
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