Using Velocity vs Time Graphs to Find Acceleration
A graph titled velocity versus time has horizontal axis time (seconds) and vertical axis velocity (meters per second). A line has 4 straight segments. Line segment A runs from 0 seconds 0 meters per second to 1 seconds 15 meters per second. Then segment B runs to 2 seconds 20 meters per second. Then segment C runs to 4 seconds 20 meters per second. Then segment D runs to 5 seconds 0 meters per second.

The acceleration of segment D is m/s2.



Rank segments A, B, and C from least acceleration to greatest acceleration.

Least:








Greatest:

Answer: Number 6 is Periods

Explanation:

Answer:

yeet yeet yeet yeet

Explanation:

Kinetic energy (K.E):-

So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.

A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.

What is the squirrel’s mass?

Answer: 0.51 kg

Answer:

either its 12 or 0.402

Explanation:

Answer:

C

Explanation:

???

i think

The question is not complete. The complete question is :

During a certain period of time, the angular position of a rotating object is given by [tex]$\theta =2t^2 +10t+5$[/tex], where θ is in radians and t is in seconds.  Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.

Solution :

Given :

Displacement or angular position of the object, [tex]$\theta =2t^2 +10t+5$[/tex]

∴ Angular speed is   [tex]$\omega = \frac{d \theta}{dt}$[/tex]

                                 ω = 10 + 4t

And angular acceleration is [tex]$\alpha = \frac{d \omega}{dt}$[/tex]

                                              α = 4

a). At time, t = 0.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(0)^2 +10(0)+5$[/tex]

                                               = 5 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(0)

                                     = 10 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

b). At time, t = 3.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(3)^2 +10(3)+5$[/tex]

                                               = 53 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(3)

                                     = 22 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

Answer:

Dimension.

Explanation:

Answer:

This question is incomplete

Explanation:

This question is incomplete but the missing figure is in the attachment below.

When an object is travelling around a circular path, there is a force that tends to draw that object towards the center of the circular path and keep the object moving in the curved path, that force is called the centripetal force. From this description, it can be deduced that the direction of the centripetal force that acts on the car (in the attachment below) is D.

The name of the force that provides this centripetal force is frictional force. This is the force that prevents the car from slipping off the road; keeping it moving in the curved path.

The energy type at the initial time is potential energy and the energy at the final time or position is kinetic energy.

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

Based on the law of conservation of mechanical energy, the formula for the change in the kinetic energy and the potential energy of the bicyclist is given as;

K.Ei + P.Ei = K.Ef + P.Ef

where;

The kinetic energy of the bicyclist increases with increase in the velocity of the bicyclist while the potential energy increases with increase in the height of the bicyclist.

At the initial position when the bicyclist is at rest, the kinetic energy is zero, so the only energy at the initial position is potential energy because the height is maximum.

In addition, at the final position, the velocity of the bicyclist is maximum and the height is zero, so the only energy at the final position is kinetic energy.

Learn more about conservation of energy here: https://brainly.com/question/166559

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