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4 years ago

In a concave mirror parallel rays falling on it convergs at

Physics

1 answer:

ella [17]4 years ago

5 0

Answer:

1) In a concave mirror parallel rays falling on it converges at F and 2F.

Explanation:

Spherical mirrors can be used for magnification of images. There are basically two types of spherical mirrors and they are converging mirror and diverging mirrors. The converging mirrors are also termed as concave mirrors and its basic work is to converge or combine light rays coming from a larger distance to a single point. Mostly the light beams falling parallel to the principle axis of the concave mirror will be acting as parallel rays. And when these parallel rays fall on the mirror, the converging point can be the focal point of the mirror.

Thus the location of converging point in concave mirrors will be based on the position or distance of object from the mirror. If the object distance is very far from the twice the focal length distance of mirror, then the converging point will be the focal point or F. And if the object is placed slightly greater than twice the distance of focal point, then the image will be obtained at 2F. But the parallel beams will be converging at F and 2F.

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Answer:

Explanation:

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Before we can get the transverse speed, we need to get the frequency and the wavelength.

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Given period = 2/15 s

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$v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s$

Hence, the transverse speed at that point is 15m/s

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3 years ago

A lightweight object and a very heavy object are sliding with equal speeds along a level frictionless surface. They both slide u

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Answer:

No object rises a greater height. They both rise the same height.

Explanation:

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1/2mv² = mgh

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Since the speed is the same for both objects, the height moved is the same since the expression for the height moved is independent of their masses but dependent on their speeds.

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4 years ago

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

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Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

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3 years ago

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Answer:

It is true

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3 years ago

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Answer:

Option 2

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3 years ago

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