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Katen [24]
2 years ago
9

Since the coefficient of friction is less than 1, what does that mean about the normal force and the force of friction?

Physics
1 answer:
ludmilkaskok [199]2 years ago
7 0

Answer:

there is friction between the two things

Explanation:

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A particle that has mass m and charge q enters a uniform magnetic field that has magnitude B and is directed along the x axis. T
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Answer:

a) The trajectory will be a helical path.

b) θ = 2*π rad

Explanation:

a) Since the initial velocity of the particle has a component parallel (x-component) to the magnetic  field B , the trajectory will be a helical path.

b) Given

t = 2*π*m/(q*B)

We can use the equation

θ = ω*Δt

where

θ is the angular displacement

ω is the angular speed, which is obtained as follows:

ω = q*B/m

then we have

θ = (q*B/m)*2*π*m/(q*B)

⇒  θ = 2*π rad

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A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee
Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2}  = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\

<u>v = 7.69 x 10³ m/s</u>

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v =  \frac{2\pi r}{T}\\\\

where,

T = Time Period of Satellite = ?

Therefore,

T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\

<u>T = 5500 s = 91.67 min = 1.53 h</u>

7 0
3 years ago
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