In a battery powered flashlight, how is energy converted

Physics

1 answer:

Nataly [62]3 years ago

5 0

Answer:

The correct answers are "chemical energy into electrical energy" and then "the electrical energy into light energy". Explanation: In the battery-powered flashlight, the battery supplies the chemical energy which makes the electrons to flow in the circuit and constitutes the current

Explanation:

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An arrow is shot at an angle 38 ◦ with the horizontal. It has a velocity of 46 m/s. How high will the arrow go?

Klio2033 [76]

Answer:

40·919 m

Explanation:

Initial velocity of the arrow = 46 m/s

Angle at which it is thrown from horizontal = 38°

<h3>At the maximum height, the vertical component of velocity will be 0</h3>

Initial velocity in vertical direction = 46 × sin(38) = 28·32 m/s

From the formula

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Considering the formula in vertical direction and taking upward direction as positive

v = 0

u = 28·32 m/s

a = - g = - 9·8 m/s²

Let s be the maximum height

- 28·32² = - 2 × 9.8 × s

⇒ s = 40·919 m

∴ The arrow will go 40·919 m high

8 0

3 years ago

high school physics, no need detail explain, just give the answer, but you have to make sure thank you

andrey2020 [161]

Answer:

approximately 30 degrees

Explanation:

If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component $(v_y)$ of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):

$v_f=v_i-a\,t\\v_f=v_y-g\,t\\0=v_y-9.8\,*\,2\\v_y=9.8\,*\,2=19.6 \frac{m}{s}$

By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:

$sin(\theta)=\frac{opp}{hyp} \\sin(\theta)=\frac{19.6}{40}\\\theta=arcsin(\frac{19.6}{40})\\\theta=29.34^o$