A car travels 150 meters East of its original starting position in 15s. What is the cars velocity?

A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionle

Radda [10]

Answer:

a) x=0 %T=0, b) x= A %T=100%, c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

x = A cos (wt + Ф)

where angular velocity is related to frequency and period

w = 2π f = 2π / T

we can write the equation of the oscillation

x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

% T = 0

b) for x = A

let's find the angle

cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

% T = 100%

c) the point of maximum compression x = -A

let's look for the angles

cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

% T = 50%

6 0

2 years ago

A family is skating at an ice rink. The 58.2 kg mother is holding the

MariettaO [177]

Answer:

When I got this question I had to draw it out so if you have to do that, draw 3 stick figures holding hands, one representing the mother, father, and daughter. Then you write their weights on top of them and then draw an arrow pointing from the father to the mother.

Explanation:

use this formula :

$a_{y}$ = $\frac{Fdadshandy}{msys}$

then you fill it in :

$a_{y}$ = $\frac{100N}{35.5kg+58.2kg}$

$a_{y}$ = $\frac{100N}{93.7kg}$

$a_{y}$ = $1.0672$ $m/s^{2}$

then you multiply that with the daughters weight :

$T_{2} x= m_{2} a_{y}$

$T_{2} x = 35.5kg (1.0672 m/s^{2})$

$T_{2} x = 37.89N$

and that's the answer :) : 37.89N

5 0

3 years ago

A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?

Andrew [12]

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
   (15 m/s) in the direction opposite to the direction you call positive.

Displacement = (distance between start-point and end-point)
   in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
   = (1/2) (3 m/s²) (25 s²) = 37.5 meters

Displacement =
   37.5 meters in the direction opposite to the direction you call positive.

5 0

3 years ago

3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a

inessss [21]

Answer:

The average angular acceleration is $\alpha =125.487 rad /s^2$