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4 years ago

11

An electron in an x-ray machine is accelerated through a potential difference of 1.00 × 104 v before it hits the target. What is

the kinetic energy of the electron in electron volts?

1 answer:

Phoenix [80]4 years ago

6 0

Kinetic energy of electron is gained by the potential energy that it is getting through the potential applied across it

now here we know that that

$V = 1.00 \times 10^4 Volts$

charge that is accelerated is an electron

so we will have

$U = qV$

here we know that

$q = 1e$

$U = (e)(1.00 \times 10^4 V)$

$U = 1.00 \times 10^4 eV$

so kinetic energy will be same as potential energy which is equal to

$KE = 1.00 \times 10^4 eV$

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Example of items that changed chemically

Ostrovityanka [42]

Burning of gases is one the example of chemical change

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4 years ago

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon’

Yuki888 [10]

Answer:

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

$s = ut + \frac{1}{2}at^2$

where u = initial velocity = 0 m/s

a = acceleration = $250 m/s^2$

t = time = 0.02 s

Therefore:

$s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m$

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

$distance = speed * time$

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

$v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s$

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

0.05 + 0.15 = 0.2 m

8 0

3 years ago

A boy drops a 0.10 kg stone down a 150 m well and listens for the echo. The air temperature is 20°C. How long after the stone is

S_A_V [24]

Use formula for Echo which is Velocity=2(Distance)/time so 343=2(150)/T 343T=300..T=300/343=0.9 seconds

4 0

3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

A graph of x vs. t2 is linear, and intercepts the vertical axis at 12 m and the horizontal axis at 4 s2. What is the function?

gogolik [260]

Answer:

x = -3t² + 12

Explanation:

x vs t² is a line.

x = at² + b

The y intercept is 12.

x = at² + 12

At t² = 4s², x = 0.

0 = a(4) + 12

a = -3

Therefore, the function is:

x = -3t² + 12

4 0

4 years ago