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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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Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous
Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

d_2 = 0.86d_1

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

Thus, speed in smaller section is 48.6 m/s

3 0
2 years ago
A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di
BlackZzzverrR [31]

Answer:

r =3 *\sqrt{2} = 4.24 cm

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

E_{MAX}= \frac{KQ}{r^{2}}

electric field inside the sphere can be determine by using below relation

\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}

r = \sqrt{2}R

r =3 *\sqrt{2} = 4.24 cm

4 0
2 years ago
Whipple is confused about the connection between the velocity and acceleration of the tennis ball. he decides to compare the vel
tamaranim1 [39]

The speed of the ball is always zero and the acceleration is always -g when it reaches the top of its motion. This is because when the ball is free, only gravity acts on it which is always downwards, hence g is the net acceleration and it is always negative. However the velocity does not direction change instantly, negative acceleration first slows down the ball with a positive velocity, until that point the ball keeps moving up, then the ball velocity becomes zero just before changing direction and becoming negative after which the ball will now go down along gravity. Hence the ball velocity is zero at the top (neither going up nor down). Mathematically this can be seen as velocity is the integration of acceleration.

7 0
2 years ago
While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe
kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

P_{Miy} + p_{Piy} + j_{y}= P_{Mfy} +P_{pfy}

m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sinФ

60 ·8 + 0 = (60 + 80)V_{f}sinФ

480 = 140V_{f} sinФ................. (I)

y-axis component form (+y north);

P_{Mix} + p_{Pix} + j_{x} = P_{Mfx}+ P_{pfx}

m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cosФ

0 + 80.9 = (60 + 80)V_{f}cosФ

 720= 140V_{f}cosФ

140Vf=\frac{720}{cos}Ф......................................(2)

 Substituting (2) into (1) to give the angle;

 480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

 

7 0
2 years ago
Two male moose charge at each other with the same speed and meet on a icy patch of tundra. As they collide, their antlers lock t
Ksenya-84 [330]
Change in velocity of larger moose: (1/3)v - v = -(2/3)v 
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2

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4 0
2 years ago
Read 2 more answers
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