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1 year ago

What is the acceleration due to gravity on earth

Physics

2 answers:

Zanzabum1 year ago

5 0

The acceleration due to gravitation on earth is 9.806 m/s^2

Svetradugi [14.3K]1 year ago

4 0

The acceleration due to gravity on earth is 9.8 m / s^2

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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.

8 0

3 years ago

The eye of a hurricane passes over grand bahama island in a direction 60.0Â° north of west with a speed of 41.0 km/h. three hour

Ksivusya [100]

initial velocity is given as 41 km/h at 60 degree North of West

$v_i = -41 cos60\hat i + 41 sin60 \hat j$

$v_i = -20.5 \hat i + 35.5 \hat j$

After some time the velocity is given as

$v_f = 25 \hat j$

now we can find the acceleration

$a = \frac{v_f - v_i}{t}$

$a = \frac{25\hat j - (-20.5 \hat i + 35.5 \hat j)}{3}$

$a = 6.83 \hat i - 3.5 \hat j$

now the distance is given by

$d = v*t + \frac{1}{2} at^2$

$d = (-20.5 \hat i + 35.5 \hat j)*4.5 + 0.5*(6.83 \hat i - 1.5 \hat j)* 4.5^2$

$d = -92.25\hat i + 159.75\hat j + 69.15\hat i - 15.19\hat j$

$d = -23.1 \hat i + 144.56\hat j$

so the magnitude of distance is

$d = \sqrt{23.1^2 + 144.56^2} = 146.4 km$

8 0

3 years ago

Read 2 more answers

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.