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Pavlova-9 [17]
3 years ago
7

Why does the frequency of a wave increase as the wavelength decreases?

Physics
2 answers:
lubasha [3.4K]3 years ago
8 0
Correct option A

The wave speed remains constant.So if the frequency increases then wavelength increases and if frequency decreases then wavelength increases to maintain the constant wave speed.

Hope This Helps You!
Mashutka [201]3 years ago
3 0

Answer: The correct answer is " because the wave speed remains constant".

Explanation:

The expression of the wavelength in terms of the frequency is as follows;

\lambda =\frac{c}{f}

Here, c is the speed of the light, f is the frequency of the wave and \lambda is the wavelength of the wave.

The wavelength of the wave is inversely proportional to the frequency of the wave as the speed of the wave remains constant.

If the wavelength of the wave decreases then the frequency of the wave increases.

Therefore, the frequency of a wave increases as the wavelength decreases because the wave speed remains constant.

The correct option is (A).

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3 years ago
If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
SP: Calculate the moment
ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, F=80\ N

Distance of application of force from the point about which moment is needed is, d=25\ cm=\frac{25}{100}\ m=0.25\ m

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0
3 years ago
What is a gas-like mixture that is made of charged particles?<br><br> (Physical Science)
Juli2301 [7.4K]

Answer:

I believe the answer is Plasma

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3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

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

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

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