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Arada [10]
3 years ago
11

If a net horizontal force of 175 N is applied to a bike whose mass is 50 kg what acceleration is produced?

Physics
1 answer:
Crazy boy [7]3 years ago
7 0

The acceleration of the bike is 3.5 m/s^2

Explanation:

We can solve this problem by using Newton's second law of motion, which states that:

F=ma

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the bike in this problem, we have

F = 175 N is the net force

m = 50 kg is the mass

And solving for a, we find the acceleration:

a=\frac{F}{m}=\frac{175}{50}=3.5 m/s^2

Learn more  about Newton's second law and acceleration:

brainly.com/question/3820012

#LearnwithBrainly

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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>

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Every galaxy is moving away from each other - not just us. And the further they are away, the faster they are moving
4 0
3 years ago
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If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds
BabaBlast [244]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

According to above question ~

  • Current (I) = 4 Amperes

  • Time (t) = 3 seconds

  • Charge (q) = ?

Let's find the charge (q) by using formula ~

  • I =  \dfrac{q}{t}

  • 4 =  \dfrac{q}{3}

  • q = 4 \times 3

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Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

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Answer:

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Answer:

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