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horsena [70]
2 years ago
10

What is parabolic motion?​

Physics
1 answer:
Cerrena [4.2K]2 years ago
8 0

Answer:

Pro jectile motion, also known as parabolic motion, consists in launching a body with a velocity that form an angle α with the horizontal

Explanation:

~♥~

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Newton’s law of gravity says the gravitational force between two objects is proportional to:
Nat2105 [25]

Answer:

Mass of the object or earth

3 0
3 years ago
A pendulum has a 0.35\ \text{kg}0.35 kg0, point, 35, space, start text, k, g, end text mass oscillating at a small angle from a
expeople1 [14]

Answer:

The frequency of oscillation of the simple pendulum is 0.49 Hz.

Explanation:

Given that,

Mass of the simple pendulum, m = 0.35 kg

Length of the string to which it is attached, l = 1 m

We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{1}} \\\\f=0.49\ Hz

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.

4 0
3 years ago
A motor has a rotor (with armature) of moment of inertia ????m . The rotor is attached to a gear box of gear ratio G > 1. The
Inessa [10]

Answer:

hello your question is incomplete attached below is the complete question

answer : The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor  i.e. J > Jm

Explanation:

Gear ratio G > 1

a) Determine the moment of inertia felt by the motor

moment of inertia felt by Motor = moment of Inertia at the armature

b) Determine the moment of inertial felt by someone who is rotating the mass by hand

moment of inertia felt by someone is = J

The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor

attached below is a detailed solution

3 0
3 years ago
Object A is positively charged. Object A and Object B
ozzi

Answer: object B is negatively charged, object C is positively charged and object D is also positively charged

Explanation: since unlike charges attract and like charges repel, for object A which is positively charged and B to attract B must be negatively charged and then for B which is negatively charged and C to attract C must be positively charged and for C and D to repel they have to be of thesame charge which means D is positive as well.

4 0
2 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
2 years ago
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