What is the speed of a car that travels a distance of 60 meters in a time of 10 seconds

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

Even though the Sun has a greater mass than Earth, the Moon orbits Earth because it's ___________________ to Earth than to the S

Pani-rosa [81]

<span>Even though the Sun has a greater mass than Earth, the Moon orbits Earth because it's closer to the Earth than to the Sun. Because of this proximity between the Earth and the Moon, the Earth has a stronger gravitational pull than the Sun does. Furthermore, the Earth's mass is 81 times that of the Moon, and so at this proximity, it is more than able to overpower what pull the Sun exerts on the Moon.</span>

3 0

3 years ago

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Velocity vector and acceleration vector in a uniform circular motion are related as.

mr_godi [17]

They are related as $\bold{\underline {v}\,.\,\underline a }= \bold{0}$

In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
In mathematics, 2 vectors ( $\underline p$ , $\underline q$ ) that are perpendicular to each other have a quality that their dot product ( $\underline p\,.\, \underline q$ ) equal to zero vector ( $\bold 0$ ) which is written as $\undeline p\,.\, \underline q = \bold 0$ .
This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner $\underline v\,.\, \underline a =\bold 0$ .