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3 years ago

14

A car increases its speed from 9.6 meters per second to 11.2 meters per second in 4.0 seconds. The average acceleration of the c

ar during this 4.0-second interval is

2 answers:

Nataly_w [17]3 years ago

6 0

A=(vf-vi)/t
a=(11.2-9.6)/4
a=0.4m/s^2

Gwar [14]3 years ago

5 0

The acceleration is the change of speed/velocity over time. Thus to calculate this you do (V1-V2)/T or (11.2-9.6)/4 or 0.4 m/s^2

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The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane

umka21 [38]

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$ .

6 0

4 years ago

Olegator [25]

Answer:

v2^2 - v1^2 = 2 g s fundamental formula

v2 = v1 + 2 g = v1 + 19.8 increase in velocity in 2 sec

v1^2 + 39.6 v1 + 392 - v1^2 = 2 * 9.8 * 123.1 = 2412.76

v1 = (2412.76 - 392) / 39.6 = 51.03

v2 = 51.03 + 19.6 = 70.63

T = 70.63 / .8 = 7.207 sec time to fall height of tower

S = 1/2 g T^2 = 4.9 * 7.207^2 = 254.5 m

(Note v2^2 - v1^2 = 70.63^2 - 51.03^2 = 2385 m

2385 / (2 * 9.8) = 122 m (close to 123.1 as was given

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3 years ago

Which part of the electromagnetic spectrum has a lower frequency than visible light?

Mademuasel [1]

The Correct Answer is <u>D.Infrared/</u> <em>INFARED has a lower frequency than visible light/</em>

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3 years ago

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chubhunter [2.5K]

-- A tornado follows a path that's a few miles wide, for a few hours.
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3 years ago

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7. A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0

Illusion [34]

Answer:

The distance between first-order and second-order bright fringes is 12.66mm.

Explanation:

The physicist Thomas Young establishes through its double slit experiment a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

$\Lambda x = L\frac{\lambda}{d}$ (1)

Where $\Lambda x$ is the distance between two adjacent maxima, L is the distance of the screen from the slits, $\lambda$ is the wavelength and d is the separation between the slits.

The values for this particular case are:

$L = 2.0m$

$\lambda = 633nm$

$d = 0.100mm$

Notice that is necessary to express L and $\lambda$ in units of milimeters.

$L = 2.0m \cdot \frac{1000mm}{1m}$ ⇒ $2000mm$

$\lambda = 633nm \cdot \frac{1mm}{1x10^{6}nm}$ ⇒ $6.33x10^{-4}mm$

Finally, equation 1 can be used:

$\Lambda x = (2000mm)\frac{(6.33x10^{-4}mm)}{(0.100mm)}$

$\Lambda x = 12.66mm$

Hence, the distance between first-order and second-order bright fringes is 12.66mm.

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3 years ago