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Artist 52 [7]
3 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]3 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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Answer:

The acceleration of the ball is 4.18 [m/s^2]

Explanation:

By Newton's second law we can find the acceleration of the ball

F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]

Now we have:

a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]

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Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, x, of an SHM is given by

x = A\cos(\omega t)

A is the amplitude and \omega is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286

0.12\omega =\cos^{-1}0.4286

0.12\omega = 1.13

\omega = 9.4

The period, T, is related to \omega by

T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67

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B : It will avoid the distortion from the atmosphere and give scientists more accurate data

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Answer:

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Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

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5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

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the equations of motion are the same.

4 0
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