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ololo11 [35]
4 years ago
6

A space probe is orbiting a planet on a circular orbit of radius R and a speed v. The acceleration of the probe is a. Suppose ro

ckets on the probe are fired causing the probe to move to another circular orbit of radius 0.5R and speed 2v. What is the magnitude of the probe’s acceleration in the new orbit?
Physics
1 answer:
Mamont248 [21]4 years ago
7 0

Answer:

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

a_{new} = 8a.

Explanation:

given data:

radius of orbit = R

Speed pf planet = v

new radius = 0.5R

new speed = 2v

we know that acce;ration is given as

a = \frac{v^{2}}{R},

a_{new} =\frac{(2v)^{2}}{0.5R},

           = \frac{4v^{2}}{0.5R}

          = \frac{8 v^{2}}{R}

a_{new} = 8a.

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

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(a) The time taken for the projectile to reach the maximum height is 32.65 s.

(b) The horizontal range of the projectile is 9,049.1 m.

The given parameters:

  • Initial velocity of the projectile, u = 320 m/s
  • Angle of projection, = 30 degrees

The time taken for the projectile to reach the maximum height is calculated as follows;

v_f = u- gt\\\\0 = 320 - 9.8t\\\\9.8t = 320\\\\t = \frac{320}{9.8} \\\\t = 32.65 \ s

The horizontal range of the projectile is calculated as follows;

R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{320^2 \times sin(2\times 30)}{9.8} \\\\R = 9,049.1 \ m

Learn more about horizontal range here: brainly.com/question/12870645

6 0
3 years ago
After performing some calculations, your calculator gives you the following results. How will you round them before presenting t
umka2103 [35]

Answer:

⇒ 401.6 ± 7.1 ms

⇒ 57.2 ± 1.6 kg

⇒ 8.1 × 107 ± 2.9 × 105 m

⇒ 2.1 ± 8.8 µN

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Explanation:

As given all equation we do without calculation we will  get here round these calculations which is given below as  

a) 401.6473 ± 7.0912 ms

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b) 57.212 ± 1.612 kg

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c) 8.12754×107 ± 2.847×105 m

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d) 2.94 ± 8.803 µN

⇒ 2.1 ± 8.8 µN

e) 294 ± 0.00481 cm

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The compressed-air tank ab has a 250-mm outside diameter and an 8-mm wall thickness. it is fitted with a collar by which a 40-kn
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Four identical particles of mass 0.514 kg each are placed at the vertices of a 4.70 m x 4.70 m square and held there by four mas
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In order to solve the problem it is necessary to take into account the concepts related to the moment of inertia, and the center of mass of the object.

Our values are given by:

m = 0.514kg

I = 4.7m (each side)

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B) For the case when the axis passes through the midpoint of one of the sides and is perpendicular to the plane of the square. The formula is given by,

I = 2m(\frac{l}{2})^2+2m(\frac{l}{2}^2+l^2)\\I = 2*(0.514)*(\frac{4.7}{2})^2+2*(0.514)*((\frac{4.7}{2})^2+4.7^2)\\I = 34.06Kgm^2

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