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liq [111]
3 years ago
11

A 2-kg ball is thrown at 3 m/s. What is the ball's momentum? *

Physics
1 answer:
crimeas [40]3 years ago
8 0

Answer:

Given

mass (m) =2kg

velocity (v) =3m/s

momentum (p) =?

Form

p=mv

2kgx3m/s

p=6kg.m/s

the momentum of ball's =6kg.m/s

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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
3 years ago
The flow of electricity can be compared of water in
Anna007 [38]

The flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

<h3>How we compare the flow of electricity to water?</h3>

Water flowing in pipes is like flowing of electricity in a circuit. A battery is like a pump from where electricity comes and moves in the circuit. Electrons flowing through wires are like water molecules flowing through pipes. So in comparison between water and electricity, both water and electricity are similar to each other in flowing and movement.

So we can conclude that the flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

Learn more about electricity here: brainly.com/question/776932

#SPJ1

7 0
2 years ago
A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If
inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

          v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

          v² = 2 a₁ x

let's calculate

          v = \sqrt {2 \ 15.0 \ 645}

          v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

          v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

         0 = v² - 2 a₂ x₂

         x₂ = v² / 2a₂

let's calculate

         x₂ = \frac{ 143.666^2 }{2 \  18.2}

         x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0
3 years ago
If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral
leonid [27]

Answer:

\theta = tan^{-1}\mu

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

F_f = mgsin\theta

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

F_f = \mu N

N = mg cos\theta

so we have

F_f = \mu (mg cos\theta)

so we will have

mg sin\theta = \mu (mg cos\theta)

so now we have

tan\theta = \mu

so maximum possible angle of the inclined plane is

\theta = tan^{-1}\mu

3 0
3 years ago
What is the mach to term of kinetic energy
lord [1]
Kinetic energy means fast move movement
5 0
3 years ago
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