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morpeh [17]
3 years ago
9

A ball is spun around in circular motion such that its frequency is 10 Hz.

Physics
1 answer:
Scilla [17]3 years ago
5 0

Answer:

a = 0.1 s b. 10 s

Explanation:

Given that,

The frequency in circular motion, f = 10 Hz

(a) Let T is the period of itsrotation. We know that,

T = 1/f

So,

T = 1/10

= 0.1 s

(b) Frequency is number of rotations per unit time. So,

t=\dfrac{n}{f}\\\\t=\dfrac{100}{10}\\\\t=10\ s

Hence, this is the required solution.

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If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
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The correct answer is letter A. 6 millimeters. <span>If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.
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Solution:
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3 years ago
In the photoelectric effect, the greater the frequency of the illuminating light, the greater the:_______
TEA [102]

Answer:

B. Maximum velocity of ejected electrons.

Explanation:

The ejection of electrons form a metal surface when the metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength or higher frequency (or equivalently, above a threshold frequency),  which leads to the enough energy of the wave to incident and get absorbed to the exposed surface emits electrons. This phenomenon is known as the photoelectric effect or photo-emission.

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The current produced as a result is called photo electricity.

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8 0
3 years ago
Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

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or

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thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}  

Thus, the minimum distance to target = \frac{290}{2} = 145 m

6 0
3 years ago
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jeyben [28]
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