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2 years ago

10

Find the speed of a satellite in geostationary orbit.

1 answer:

mixas84 [53]2 years ago

4 0

Answer:

A geostationary orbit can be achieved only at an altitude very close to 35,786 kilometres (22,236 miles) and directly above the equator. This equates to an orbital speed of 3.07 kilometres per second (1.91 miles per second) and an orbital period of 1,436 minutes, one sidereal day

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A computational model predicts the maximum potential energy a roller coaster car can have given its mass and its speed at the lo

andreyandreev [35.5K]

Answer:

The prediction for its maximum potential energy is 109,375 J

Explanation:

Given;

mass of the coaster car, m = 350 kg

speed of the coaster car at the lowest point, v = 25 m/s

The coaster car will have maximum kinetic energy at the lowest point and based on law of conservation of mechanical energy, the maximum kinetic energy of the coaster car at the lowest point will be equal to maximum potential energy at the highest point.

$K.E_{max} = P.E_{max}$

$K.E_{max} = \frac{1}{2} mv^2\\\\K.E_{max} = \frac{1}{2} (350)(25)^2\\\\K.E_{max} =109,375 \ J$

$Thus, P.E_{max} = 109,375 \ J$

Therefore, the prediction for its maximum potential energy is 109,375 J

3 0

3 years ago

A 68 kg object, starting from rest, travels from point A to point B at a rate of 30 m/s in 2 hours. What is the applied force on

Natasha2012 [34]

Answer:

$\huge\boxed{\sf F = 0.28\ N}$

Explanation:

<h3>Given Data:</h3>

Mass = m = 68 kg

Velocity = v = 30 m/s

Time = 2 hours = 2 × 60 × 60 = 7200 s

<h3>Required:</h3>

Force = F = ?

<h3>Formula to be used:</h3>

$\displaystyle F = \frac{mv}{t}$

<h3>Solution:</h3>

$\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}$

7 0

1 year ago

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

4 years ago

To understand the formula representing a traveling electromagnetic wave.

Mumz [18]

Answer:

Explanation:

Time period is the reciprocal of frequency

T = 1/F

F = 1/T

but angular frequency w = 2πF

F = w/2π

The detailed steps is as shown in the attached file

8 0

3 years ago

After hearing about an accident on his normal route, Mr. Gujral checks for alternate routes to get to work. What type of circuit

shtirl [24]

Answer:

C

Explanation:

a parallel circuit because there is more than one path

3 0

3 years ago