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Grace [21]
3 years ago
7

An aeroplane, flying in a straight line at a constant

Physics
1 answer:
dedylja [7]3 years ago
3 0

Answer:

200

Explanation:

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a) the charge of an electron is equivalent to the magnitude of the elementary charge but barring a negative sign since the side of the elementary charge is roughly 1.602 * 10 - 19 Columbus then the charge of the electronic is-1.602 * 10 - 19

b) b=2T on the electron moving in the magnetic field

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2 years ago
Please help :)<br><br><br> Microwaves are transmitted by ....... ?
Dima020 [189]
Microwaves are transmitted by Radio Waves.
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Type O blood is known as the universal donor why is this so?
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Because their blood cells could be transfused to any blood type
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7 0
3 years ago
Read 2 more answers
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

6 0
3 years ago
A 645-turn coil with a 20.250 m ​2 ​​ area is spun in Earth’s 5.00×10 ​−5 ​​ T magnetic field, producing a 1.25-V maximum emf. A
Dmitriy789 [7]

Answer:

\omega = 1.914\ rad/s

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

\varepsilon = NAB\omega

\omega= \dfrac{\varepsilon}{NAB}

\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}

\omega = 1.914\ rad/s

Angular velocity of the coil = \omega = 1.914\ rad/s

5 0
3 years ago
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