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Artist 52 [7]
2 years ago
14

the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w

ith a mass of 3.26x10^31 kg has a kinetic energy of 1.10x10^42 j, what is its speed?
Physics
1 answer:
4vir4ik [10]2 years ago
6 0

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

                   K.E = 1/2 \times m \times v^{2}

where K.E represents the kinetic energy,

           m represents the mass,

            v represents the velocity.

                  K.E = 1/2 \times m \times v^{2}

    1.10 \times 10^42 = 1/2 \times 3.26 \times 10^31 \times v^{2}

                     v^{2} = (1.10 \times 10^42 \times 2) / (3.26 \times 10^31)

                     v = 0.956 m / s.

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A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold
salantis [7]

Explanation:

Below is an attachment containing the solution.

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2 years ago
An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init
elena-14-01-66 [18.8K]

Answer:

Approximately 71\%.

Explanation:

The formula for the kinetic energy \rm KE of an object is:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2,

where

  • m is the mass of that object, and
  • v is the speed of that object.

Important: Joule (\rm J) is the standard unit for energy. The formula for \rm KE requires two inputs: mass and speed. The standard unit of mass is \rm kg while the standard unit for speed is \rm m \cdot s^{-1}. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

m = 63\; \rm g = 0.063\; \rm kg.

Initial \rm KE of this arrow:

\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}.

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%.

8 0
2 years ago
Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre
alex41 [277]

Answer:

The simplified expression is M  =  \frac{v^2 r}{G}

Explanation:

From the question we are told that  

     M  = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }

So simplifying we have

    M  =    \frac{m v^2}{r} *  \frac{r^2 }{ mG }

    M  =  \frac{v^2 r}{G}

Thus the simplified formula is M  =  \frac{v^2 r}{G}

3 0
2 years ago
Two objects, X and Y, move toward one another and eventually collide. Object X has a mass of 2M and is moving at a speed of 2v0
Sergeu [11.5K]

Answer:

c. The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.

Explanation:

If we take both objects as one single system, during the collision, assuming no external forces acting, the only forces present are the one that object X exerts on object Y, and the force that object Y does on object X.

These two forces, according to Newton's 3rd Law, form an action-reaction pair, and consequently, are equal in magnitude, acting in opposite directions.

As the object X is moving to the right, the force that produces, (F), is in the same direction (on Y), while for object Y, moving to the left, the force that produces (F also in magnitude) is in the same direction (on X), so the right answer is c.

The effect of the forces is different, due to masses are different, according Newton's 2nd Law.

6 0
2 years ago
To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj
ExtremeBDS [4]

Answer:

1.138\times 10^{-3}V

Explanation:

Apply Faraday's Newmann Lenz law to determine the induced emf in the loop:

\epsilon=\frac{d\phi}{dt}

where:

d\Phi-variation of the magnetic flux

dt-is the variation of time

#The magnetic flux through the coil is expressed as:

\Phi=NBA \ Cos \theta

Where:

N- number of circular loops

A-is the Area of each loop(A=\pi r^2=\pi \times 5^2=78.5398)

B-is the magnetic strength of the field.

\theta=15\textdegree- is the angle between the direction of the magnetic field and the normal to the area of the coil.

\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}

\frac{dB}{dt}-=0.0250T/s is given as rate at which the magnetic field increases.

#Substitute in the emf equation:

=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V

Hence, the induced emf is 1.138\times 10^{-3}V

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