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GREYUIT [131]
3 years ago
9

Consider an isolated system, by which we mean a system free of external forces. If the sum of the particle energies in the syste

m increases by 147 joules, what must have been the change deltacapU in the system's potential energy?
Physics
1 answer:
Airida [17]3 years ago
5 0

Answer:

The system's potential energy is -147 J.

Explanation:

Given that,

Energy = 147 J

We know that,

System is isolated and it is free from external forces.

So, the work done by the external forces on the system should be equal to zero.

W=0

We need to calculate the system's potential energy

Using thermodynamics first equation

\Delta U=W-\Delta E

Put the value into the formula

\Delta U=0-147

\Delta U=-147\ J

Hence, The system's potential energy is -147 J.

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A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do
Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

\theta=90-40=50^o

From the triangle OAB

cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}

1340.57=35^2+25^2-x^2

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

4 0
3 years ago
8) Find the X and Y component of 10degree vector that has 5N.
Deffense [45]

Answer:

Fx  = 4.92 [N]

Fy = 0.868 [N]

Explanation:

Let's take the 10 degrees as a measure from the horizontal component to the vector.

Thus taking the components in the X & y axes respectively:

Fx = 5*cos(10) = 4.92 [N]

Fy = 5*sin(10) = 0.868 [N]

3 0
3 years ago
A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its
dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

             <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

         (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

              (150 million) x  (1 / 3,262,000) x (70 km/sec) =

                                 <em>3,219 km/sec  </em>in the direction away from us (rounded)

4 0
3 years ago
A space station rotates, providing 1 g acceleration for its inhabitants. If it is
disa [49]

Answer:

Hype

Explanation:

Yyrru

4 0
3 years ago
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