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JulsSmile [24]
3 years ago
9

Do you know which component of the Star Wars movies is possible according to our current understanding of physics?

Physics
1 answer:
gayaneshka [121]3 years ago
8 0
The space opera interstellar<span> epic </span>Star Wars<span> uses </span>science<span> and </span>technology<span> in its settings and storylines, although its main focus is not necessarily on science. The component of the Star Wars movies which is possible according to our current understanding of physics is their interstellar travel.
</span>The behavior of spacecraft engaged in battle is another aspect of the films is troubling the scientists. According to them,  when hit in space by a blaster, the crafts should continue moving in the same direction as the impact rather than falling downwards as the effects of gravity would not apply while in orbit. <span>


</span>
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Discuss the difference between waveform graphs and vibration graphs​
Kitty [74]

Difference exists mainly in the label for x axis.

Explanation:

  • Shapes of waveform and vibration graphs are same.
  • Vibration graphs shows the particle at a single location in the path of the wave when time passes.
  • Waveform graphs shows the particle at multiple locations at a single moment of time.
6 0
3 years ago
A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
3 years ago
A supercapacitor is an electrical energy storage device. A supercapacitor, initially charged to 2.1 thousand millivolts, supplie
svet-max [94.6K]

Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %  

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

A = P × ( (1 - r/100 )ⁿ

where A IS 0.8, P is 2.1, r is 9.9

so we substitute

0.8 = 2.1 × ( 1 - 0.099 )ⁿ

0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

Therefore, the time taken t is 9.25 minutes

6 0
3 years ago
In the simulation above, as the projectile travels upward, how does the vertical velocity change? Question 9 options:
enot [183]

Answer:

Vertical velocity decreases.

Explanation:

The motion of the ball is a projectile ball, which consists of two independent motions:

- a horizontal motion, with constant velocity

- a vertical motion, with constant acceleration g=9.8 m/s^2 towards the ground

In the vertical motion, there is a constant acceleration directed downward: this means that the vertical velocity decreases as the ball goes higher. In fact, it decreases following the equation

v(t)=v_0 -gt

And it decreases until the ball reaches its maximum height, then it starts increasing again.

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Answer:

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