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Solnce55 [7]
3 years ago
12

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the o

bject is 1.15 m/s, and its maximum acceleration is 6.52 m/s2. How much time elapses between an instant when the objects speed is at a maximum and the next instant when its acceleration is at a maximum?
Physics
1 answer:
Readme [11.4K]3 years ago
5 0
<span>Object's speed is max when passing through 0 displacement.
Object's acceleration is max at either end of its oscillation, and these two times are 1/4 period apart.
y = A sin (wt)
v= Aw cos(wt)
a = - A (w^2) sin (wt)
 
v max = Aw
 a max= -A w^2
 These relations can give you w= 2 pi/ T where T = period of oscillation.</span>
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Name at least four factors involved in performing a good throw.
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Explanation:

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4 0
3 years ago
Read 2 more answers
A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis
GarryVolchara [31]

Resultant displacement is 29.2 km at 83.1^{\circ} north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

- First displacement is 25.5 km east, therefore its components are

A_x = 25.5 km\\A_y = 0 km

- Second displacement is 41.0 km northwest, so its components are

B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km

So, the components of the resultant displacement are

R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km

And so, the magnitude is calculated using Pythagorean's theorem:

R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km

And the direction is given by

\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}

Where the angle is measured from the west direction, since Rx is negative.

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

3 0
3 years ago
1. A 65 kg student, starting from rest, slides down an 16.2 m high water slide. On the way down, friction does 5700 J of work on
GuDViN [60]

Answer:

11.94

Explanation:

Remark

Find the Potential Energy at the top.

Givens

m = 65 kg

h = 16.2 m

g = 9.81

PE = 65 * 9.81 * 16.2

PE = 10329.93

The tricky part is what do you do about Friction?

Formula

PE = Friction + KE

Solution

PE = 10329.93 Joules

Friction = 5700 Joules

Find the KE

10329.93 = 5700 + KE

KE = 10329.93 - 5700

KE = 4629.93

Find V from the KE formula

KE = 4629.93

m = 65

KE = 1/2 m v^2

KE = 1/2 65 v^2

4629.93 = 1/2 65 v^2

v^2 = 142.46

v = √142.46

v = 11.94

4 0
3 years ago
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