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likoan [24]
3 years ago
15

This means that the speed at which the bullet travels across Earth's surface (its magnitude of horizontal velocity) does not aff

ect _____.
the speed at which it falls toward the Earth

the rate at which it slows down

the distance it will travel

friction from the air
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0

Answer: the speed at which it falls toward the Earth.


Explanation:


A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.


Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).


This  kind of motion is the result of two independent motions: vertical motion and horizontal motion.


The observed net velocity is the vectorial sum of the vertical and horizontal velocities.


The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).


The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.


Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.

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Naddik [55]

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

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The centripetal acceleration is given by

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For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

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Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

    G m M / r² = m a

    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

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Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

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      r = 0.3787 10¹² m

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7 0
3 years ago
how fast would a(n) 73 kgkg man need to run in order to have the same kinetic energy as an 8.0 gg bullet fired at 430 m/sm/s ?
Mars2501 [29]

Answer:

an 85 kg person run to equal the kinetic energy of an 8.0 g bullet fired at 410 m/s? The speed is, it might be argued,

v=4.0m/s

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KE=\frac{1}{2} *m*v^{2}

K.E= kinetic energy

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V= speed

so,

KE=1/2*0.008*(420)*420

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v*v=672.4/85

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v=4.0m/s

Therefore

v=4.0m/s

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The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

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8 0
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10. How is the mass number calculated for an element?
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5 0
3 years ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

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Initial Kinetic energy

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= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

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d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

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= .5 x 1400 x 9.8

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So this is not possible.

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3 years ago
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irina1246 [14]

di means two

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4 years ago
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