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Liula [17]
3 years ago
7

An object of mass m is attached to a vertically oriented spring. The object is pulled a short distance below its equilibrium pos

ition and released from rest. Set the origin of the coordinate system at the equilibrium position of the object and choose upward as the positive direction. Assume air resistance is so small that it can be ignored.
Physics
1 answer:
Airida [17]3 years ago
3 0

Answer:

The motion is described by a set of explanations below

Explanation:

For the part A, the position of the moment is represented by the plot on each side in the far left. Because the mass has been pulled down, this can only assume the negative downward direction. Thus, graphs D and H will be negative on the far left at a position when the time t = 0.

For the part B, the position of the mass will be at point E.

For the part C, the mass will be at the position F. This is because the acceleration is a derivative of the velocity of the object as given by the following derivative equation:

a = \frac{d}{dt}(v)

where a = acceleration and v is the velocity.

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An object has a position given by the radius vector r = [2.0 m + (3.00 m/s)t](i)+ [3.0 m - (2.00 m/s^2)t^2](j). Here (i) and (j)
jenyasd209 [6]

Answer:

The speed of the object is (3i - 4.00tj)m/s

The magnitude of the acceleration is 4.00m/s²

Explanation:

Given - position vector;

r = (2.0 + 3.00t)i + (3.0 - 2.00t²)j       -------------------(i)

To get the speed vector (v), take the first derivative of equation (i) with respect to time t as follows;

v = \frac{dr}{dt}

 v = \frac{d[(2.0 + 3.00t)i + (3.0 - 2.00t^2)j]  }{dt}  

v  = 3i - 4.00tj      ------------------------(ii)

To get the acceleration vector (a), take the first derivative of the speed vector in equation(ii) as follows;

a = \frac{dv}{dt}

a = \frac{d(3i - 4.00tj)}{dt}

a = -4.00j

The magnitude of the acceleration |a| is therefore given by

|a| = |-4.00|

|a| = 4.00 m/s²

In conclusion;

the speed of the object is (3i - 4.00tj)m/s

the magnitude of the acceleration is 4.00m/s²

3 0
3 years ago
A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01x10⁵ Pa.
Cerrena [4.2K]

Answer:

\frac{1.019}{1}

Explanation:

To solve this equation we will have to consider that the bubble is filled with an Ideal Gas and as such we can use the Ideal Gas Law

PV = nRT

Where

P = Pressure

V = Volume

n = Moles

R = Ideal Gas Constant

T  = Temperature

Now since we know that the value for the temperature and moles is constant we can simply use Boyles Law for the two states

P_{1} V_{1} =P_{2} V_{2}

Let us look at the two states

State 1 (at top)

Pressure = 1.01*10^5

Volume = V_{1}

State 2 (at bottom)

Pressure = 1.01*10^5 + dgh

Where

d = Density of liquid (1000 kg/m³)

d = Acceleration due to gravity (9.8 m/s²)

d = Height of liquid (0.200 m)

Pressure = 102,962

Volume = V_{2}

Inputting these values into the Boyles Law

P_{1} V_{1} =P_{2} V_{2}\\ (101000)V_{1} = (102962)V_{2}\\ \frac{V_{1}}{V_{2}} = \frac{102962}{101000} \\  \frac{V_{1}}{V_{2}} = \frac{1.019}{1}

6 0
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8 0
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When are ionic bonds formed
allochka39001 [22]

Answer:

ionic bonds formed from the electrostatic attraction between oppositely charged ions in a chemical compound.

Explanation:

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How can I make the atomic model of boron?
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This should help

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8 0
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