A 0.36 kg object, attached to a spring with constant k=10n/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude 0.082 m. What is it speed when it’s displacement is 0.041 m

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

Explanation:

First, we need to determine the angular frequency of the system (), in radians per second:

(1)

Where:

- Spring constant, in newtons per meter.

- Mass, in kilograms.

If we know that and , then the angular frequency of the system is:

The kinematic formulas for the position (), in meters, velocity (), in meters per second, and acceleration of the object (), in meters per square second, are:

(2)

(3)

(4)

Where is the amplitude of the motion, in meters.

From (2) we determine the time associated with position (, ):

(5)

And the speed of the object is:

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz