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DanielleElmas [232]
3 years ago
15

An object has a charge of 4.5 c is it negative or positive

Physics
2 answers:
Dmitry [639]3 years ago
7 0
This is a positive charge.
Westkost [7]3 years ago
5 0
It has a positive charge
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3 years ago
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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

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

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

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

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's
Dmitrij [34]

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

4 0
3 years ago
Si la fuerza de repulsión entre dos cargas es 18 × 1013
sattari [20]

Answer:

Explanation:

F = kQq/r²

r = √(kQq/F)

a)  r = √(8.899(10⁹)(8)(4) / 18(10¹³)) = 0.0397749... m

  r = 40 mm

b) r = √(8.899(10⁹)(12)(3) / 18(10¹³)) = 0.0421876... m

   r = 42 mm

5 0
3 years ago
An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta
icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

v=\sqrt{\frac{T}{\mu} }

where T= 200 N is tension in the string , \mu=1.0 g/m is the linear mass density

v=\sqrt{\frac{200}{1\times 10^{-3} }

v=447.2 m/s

Wavelength of the wave in the string is

\lambda =2L=2\times 0.8=1.6 m

The frequency is

f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m

8 0
3 years ago
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