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Juliette [100K]
3 years ago
7

What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface

Physics
1 answer:
juin [17]3 years ago
3 0

Answer:

7,790.38 m/s

Explanation:

Given the following :

What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface

Altitude = 203 km

Using the formula :

V = √GM/r

Where G = gravitational constant =6.67×10^-11

Kindly check attached picture for detailed explanation.

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A ball is Thrown Upward with An initial velocity of 30 m/s , How high does it rise from the ground when the ball reaches the hig
kobusy [5.1K]

Answer:

  • The maximum height reached by the ball is 45.92 m
  • Time taken to fall down to half of its height is 2.2 s

Explanation:

Given;

initial velocity of the ball, u = 30 m/s

final velocity of the ball at the highest point, v = 0

The maximum height reached by the ball is calculated as;

v² = u² - 2gh

where;

h is the maximum height reached by the ball

0 = 30² - (2 x 9.8)h

19.6h = 900

h = 900 / 19.6

h = 45.92 m

Time taken to fall to half of its height is calculated as;

when falling down, the final velocity v becomes the initial velocity = 0.

Apply the following kinematic equation;

h = ut + ¹/₂gt²

h = 0 + ¹/₂gt²

h = ¹/₂gt²

where;

h = 45.92 m is the maximum height reached

half of h = 45.92 / 2 = 22.96 m

22.96 = ¹/₂gt²

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 22.96}{9.8} }\\\\t = 2.2 \ s

3 0
3 years ago
A 65 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5 m/s. how fast is he going as he l
KIM [24]

m = mass of trampoline artist = 65 kg

v₀ = initial speed of the artist at the top of platform = 5 \frac{m }{s}

h = height through which the artist drop before landing on trampoline = 3 m

v = final speed of the artist just before landing on trampoline = ?

using conservation of energy

Kinetic energy of artist just before landing = initial kinetic energy at the top of platform + potential energy at the top of platform

(0.5) m v² = (0.5) m v₀² + mgh

dividing each term by "m"

(0.5)  v² = (0.5) v₀² + gh

inserting the values

(0.5)  v² = (0.5) (5)² + (9.8)(3)

v = 9.2 \frac{m }{s}

x = compression of the trampoline = ?

k = spring stiffness constant = 62000 \frac{N }{m}

Assuming the lowest depression point as reference line for measuring the potential energy

using conservation of energy

kinetic energy of artist + potential energy of artist before landing = spring potential energy of trampoline

(0.5) m v² + mg x = (0.5) k x²

inserting the values

(0.5) (65) (9.2)² + (65 x 9.8) x = (0.5) (62000) x²

x = 0.31 m

6 0
3 years ago
If you start rolling down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill ypu KE
Serjik [45]

Answer:

v=\sqrt{(2x/m)}\  \ m/s

Explanation:

Potential Energy= Kinetic Energy

Let x be the value of Kinetic Energy.

We know that

PE=KE=x\\x=\frac{1}{2}mv^2\\

Make v the subject of the formula to get speed at the bottom of the hill.

v^2=2x/m\\v=\sqrt{2x/m}

8 0
3 years ago
I need a diagram for how a scrap heap magnet works
cupoosta [38]
Will this one work?...................

5 0
3 years ago
What is the current in the 20.0 resistor?
Charra [1.4K]
I=120 V/20 ohms
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SO the answer would be <span>A. 6.00 A</span>
8 0
3 years ago
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