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denpristay [2]
3 years ago
15

An object is in simple harmonic motion. Find an equation for the motion given that the frequency is 3⁄π and at time t = 0, y = 0

, and y' = 6. What is the equation of motion?
Physics
2 answers:
alexdok [17]3 years ago
7 0

Answer: y(t)= 1/π^2 sin(6*π^2*t)

Explanation:

because im smrt

Delicious77 [7]3 years ago
3 0

Answer: y(t)= 1/π^2 sin(6*π^2*t)

Explanation: In order to solve this problem we have to consider the general expression for a harmonic movement given by:

y(t)= A*sin (ω*t +φo) where ω is the angular frequency. A is the amplitude.

The data are: ν= 3π; y(t=0)=0 and y'(0)=6.

Firstly we know that 2πν=ω then ω=6*π^2

Then, we have y(0)=0=A*sin (6*π^2*0+φo)= A sin (φo)=0 then φo=0

Besides y'(t)=6*π^2*A*cos (6*π^2*t)

y'(0)=6=6*π^2*A*cos (6*π^2*0)

6=6*π^2*A then A= 1/π^2

Finally the equation is:

y(t)= 1/π^2 sin(6*π^2*t)

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If the solar system shrank so that the sun were located just one centimeter from Earth, about how far away could you find the ne
DedPeter [7]
E                  S                                                               *

The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).

The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years

Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.

To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.

As a result, you get a mere 268335.7 centimeters.

To put that into perspective, that's only about 1.7 miles

A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
8 0
3 years ago
When do phase transitions occur in molecules?
jekas [21]
It occurs when energy is supplied or withdrawn :)
5 0
3 years ago
A 2000kg suv accelerates from rest at a rate of 3.00m/s^2. The total amount of force resisting its motion 1500N. How much force
choli [55]

The total force that the SUV exerts is:

F = 2000 kg * 3 m/s^2

F = 6000 N

 

Since a resisting force amounting to 1500 N is exerted, then the force exerted by the SUV tires is:

F tire = 6000 N – 1500 N

F tire = 4500 N

7 0
3 years ago
A object weighing 5 kg, starts to accelerate evenly on a horizontal line. A force moves the object
hammer [34]
Work= force*distance
Work= x*12
Force= mass*acceleration
Force= 5 kg*6
Force= 40 N
Work= 40×12
Work= 480 J (joules)
I think this is it
5 0
4 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
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