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3 years ago

Which of the following stars is likely to emit the most energy into space?

Physics

2 answers:

alisha [4.7K]3 years ago

8 0

Answer: The brightest star

Explanation: The brightest start gives out more energy

Mariana [72]3 years ago

7 0

Answer: the brightest star

Explanation:

The more the surface area the brighter the star.

That increased surface area allows more light and energy to be given off

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What is the sequence of energy transformations from the moment the ball is dropped to the moment the board is bent to its maximu

Mila [183]

Answer:

Explanation:

Transformation of energy involves conversion of energy from one form to another for example our movement around involves the conversion of chemical energy stored in the food we eat to other forms of energy such as kinetic energy for the movement, electrical energy in the neurons for impulses and others

The ball posses gravitational potential energy since it is held at a displacement to the ground ( zero point) and when released, the gravitational potential energy is converted to kinetic energy which leads to the fall of the ball until it is at zero displacement to the earth. The board likewise when bent to its maximum extent stored elastic potential energy as a result of the partial displacement of its constituent particle provided it is not stretch beyond its elastic limit which can lead to deformation of the board and the elastic potential energy lost.

5 0

4 years ago

Winds blowing toward the east are called easterlies. True or False

eimsori [14]

Answer: False

Explanation:

Winds are named for the cardinal direction they blow from. Hence, a wind that "blows towards the east", logically should come from the west and is called a "west wind".

In thise sense, one of the best examples of this type of wind are the Westerlies, which are are prevailing winds that blow from the west at midlatitudes and have the characteristic that are stronger during winter and weaker during summer.

Therefore, the statement is false.

6 0

3 years ago

For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.