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Reil [10]
3 years ago
6

How fast would the International Space Station (ISS) have to travel to maintain a circular orbit a distance of 1400 km above the

earth?
Physics
1 answer:
professor190 [17]3 years ago
4 0

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by  v=\sqrt{\frac{GM}{r}}, where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

              v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s

  The International Space Station move at 7.22 km/s.      

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Question C) needs to be answered, please help (physics)
Zarrin [17]

(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s)   ==>   <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

3 0
3 years ago
A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
A student practicing for track ran 800 meter in 110 seconds. what was her speed?
adoni [48]
Her speed was 7.27 meters per second
3 0
3 years ago
Find the velocity. 10 points. Will give brainliest!
BARSIC [14]

Answer:

6.060606...

Explanation:

To figure out velocity, you divide the distance by the time it takes to travel that same distance, then you add your direction to it. So the distance would be 1000m and the time would be 2 minutes and 45 seconds and if you convert the minutes into fractions you would get 165 seconds than you would divide 1000m by 165 seconds and you would get 6.060606... seconds as her average velocity

4 0
3 years ago
Part B
Katena32 [7]

Answer:

While slavery was the major issue separating the North and South, it was not slavery itself that sparked the conflict. The South wanted to secede from the Union, and the North refused. While President Abraham Lincoln personally opposed slavery, he recognized that it was legal under the U.S. Constitution at the time. He also recognized that few in the North were ready to go to war to free the slaves. For Lincoln and the northern majority, preservation of the Union was the foremost goal.

4 0
3 years ago
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