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2 years ago

15

In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width

of the central bright fringe

1 answer:

ArbitrLikvidat [17]2 years ago

6 0

Answer:

It becomes wider

Explanation:

Because The bigger the object the wave interacts with, the more spread there is in the interference pattern. Decreasing the size of the opening increases the spread in the pattern.

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A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?

yanalaym [24]

Answer:

$KE=12,500J$

Explanation:

The formula for kinetic energy is:

$KE = \frac{1}{2}mv^2$

We can plug in the given values into the equation:

$KE = \frac{1}{2}*1000kg*(5m/s)^2$

$KE = 500kg*25m^2/s^2$

$KE=12,500J$

7 0

2 years ago

How to find the time with only the distance or height from the ground

Harman [31]

You use acceleration due to gravity
and 1/2 atsqr=d
therefore 1/2 * 9.8 * tsqr= d

5 0

3 years ago

A ball is dropped from the top of the building.it initially moves at 4.0 m/s after 0.5 seconds it moves at 3.8m/s what force is

Ivahew [28]

The correct answer is

Air resistance

In fact, when a ball is in free fall, there are two forces acting on it:

- its weight (force of gravity), acting downward

- the air resistance, acting upward

The effect of the weight is to accelerate the ball, because its direction is the same as the direction of motion of the ball, while the effect of the air resistance is to slow down the ball, because its direction is opposite to that of the motion.

6 0

3 years ago

If a 60 kg woman, runs up a flight of stairs having a total rise of 4.0m in a time of 4.2s. What average power did she supply?

LenKa [72]

<span>PE = m * g * h
PE = 60 * 9.8 * 4
PE = 2352 J

Power = E/t

Power = PE/t
Power = 2352/4.2
Power = 560 Watt</span>

6 0

3 years ago

A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th

timama [110]

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

$\Delta E = K_{1}-K_{2}$ (1)

Where:

$\Delta E$ - Energy losses, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

$K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2}$ (3)

$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

$\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{1} \approx 1.825\,\frac{rad}{s}$

$\omega_{1}\approx 0.291\,\frac{rev}{s}$

$K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2}$ (4)

$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

$\omega_{2} \approx 0.205\,\frac{rev}{s}$

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0

3 years ago