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2 years ago

Find the speed of a satellite in geostationary orbit.

Physics

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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Solve the equation below for vi.<br> d=vit +.5at2

Andrej [43]

Answer: $\dfrac{d-0.5at^2}{t}=v_i$

<u>Explanation:</u>

$d=v_it+0.5at^2\\\\\\\text{Subtract}\ 0.5at^2\ \text {from both sides:}\\d-0.5at^2=v_it\\\\\\\text{Divide both sides by t:}\\\dfrac{d-0.5at^2}{t}=v_i$

8 0

3 years ago

An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide

Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is $4.2\ cm$

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

$v=\sqrt{\dfrac{2eV}{m}}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}$

$v=15689290.81\ m/s$

$v=1.56\times10^{7}\ m/s$

(b). We need to calculate the radius of electron

Using formula of centripetal force

$\dfrac{mv^2}{r}=qvB$

$r=\dfrac{mv}{qB}$

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

$r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}$

$r=0.042\ m$

$r=4.2\ cm$

Hence, (a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is 4.2 cm

8 0

3 years ago

The water waves below are traveling along the surface of the ocean at a speed of 2.5 m/s and

gavmur [86]

Answer:2 seconds

Time= Distance/speed

=50/25

=2seconds

3 0

3 years ago

A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.6 m/s. The coefficient of friction betwe

USPshnik [31]

Answer:

434.16 Joules

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of person

From work-energy theorem

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}\times 67\times (3.6^2-0)\\\Rightarrow KE=434.16\ Joules$

The runner loses 434.16 Joules of mechanical energy.

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow \mu mg=\frac{W}{s}\\\Rightarrow s=\frac{W}{\mu mg}\\\Rightarrow s=\frac{434.16}{0.7\times 67\times 9.81}\\\Rightarrow s=0.94364\ m$

He slides 0.94364 m

4 0

3 years ago

Match the following items.

trasher [3.6K]

Number two is the Alpha particle

Number one is the Gamma-ray

Number three is the Beta particle

Hope this helped.

5 0

3 years ago

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