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Tanzania [10]
2 years ago
5

A meteoroid, heading straight for Earth, has a speed of 14.8 km/s relative to the center of Earth as it crosses our moon's orbit

, a distance of 3.84 × 108 m from the earth's center. What is the meteroid’s speed as it hits the earth? You can neglect the effects of the moon, Earth’s atmosphere, and any motion of the earth. (G = 6.67 × 10 -11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg)
Physics
1 answer:
Vera_Pavlovna [14]2 years ago
3 0

Answer:

v = 112.424 km / s

Explanation:

Given:

d = 3.84 x 10 ⁸ m  , R = 63 x 10³ m  , m earth = 5.97 x 10²⁴ kg  ,  G = 6.67 x 10⁻¹¹ N * m² / kg² , v₁ = 14.8 x 10 ³ m / s

Using the equation

KE + PE = initial KE + PE

¹/₂ * m * v₂² - G * m/ R = ¹/₂ * m * v₁² - G * m/(R+d)

v₂² = v₁² + 2*G * m [ 1/R - 1/(R+d) ]

v₂² = 14.8 x 10 ³ m /s + 2 * 6.67 x 10⁻¹¹ N * m² / kg² * 5.97 x 10²⁴ kg  [ 1 / 63 x 10³m - 1 / ( 63 x  10³ + 3.84 x 10⁸ m ) ]

v = √ 1.26 x 10¹⁰ m² / s²

v =  112.424 x 10³ m/s  

v = 112.424 km / s

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2 years ago
Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.
Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let v_1 and v_2 be the speed of the ball1 and the ball2 respectively, such that

v_1>v_2\;\cdots(i)

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, v, is \frac 12 m v^2

and the potential energy is due to the change in height is mgh [where g is the acceleration due to gravity]

So, the total energy of ball1,

=\frac 12 m v_1^2 + mgh\;\cdots(ii)

and the total energy of ball1,

=\frac 12 m v_2^2 + mgh\;\cdots(iii).

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, \frac 12 m v_1^2 >\frac 12 m v_2^2

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

6 0
2 years ago
Two cars collide at an intersection. One car has a mass of 1300 kg and is
Nata [24]

Answer:B

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3 years ago
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You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
Monica [59]

a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.

from above statement we got

height = 78.4 m

since the ball is thrown, so its vertical velocity would be zero

u = 0

taking g = 9.8m/s^2

now, using the equation of motion

h = ut + gt^2/2

now putting all the values in it

we got ,

78.4 = 9.8 * t^2/ 2

by solving we got,

t = 4 sec

b) now, since along the horizontal , no force acting and accelaration is zero so

R = ut , R is RANGE

R = 5 * 4

range =  20 m

c)  vertical components of the stone’s velocity just before it hits the ground = v sin θ =

horizontal   components of the stone’s velocity  just before it hits the ground = v cos θ

To know more about velocity  visit :

brainly.com/question/18084516

#SPJ9

4 0
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