Answer:
The frequency would double.
Explanation:
Given:
Speed of wave (v) = constant.
Frequency of wave initially (f₁) = 2 Hz
Initial wavelength of the wave (λ₁) = 1 m
Final wavelength of the wave (λ₂) = 0.5 m
Final frequency of the wave (f₂) = ?
We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.
Therefore, framing in equation form, we have:
Wavelength × Frequency = Speed
It is given that speed of the wave remains the same. So, the product must always be a constant.
Therefore,
Now, plug in the given values and solve for 'f₂'. This gives,
Therefore, the final frequency is 4 Hz which is double of the initial frequency.
f₂ = 2f₁ = 2 × 2 = 4 Hz
So, the second option is correct.
To calculate the force of impact F, first lets calculate the acceleration a of the ball:
a=v/t where v is the velocity of the ball and t is time
a=32/0.8=40 m/s²
To get the force F we need the Newtons second law:
F=m*a where m is the mass of the ball and a is the acceleration.
F=m*a= 0.2*40 = 8 N
So the impact force is F= 8 N.
A binomial nomenclature, more commonly referred to as a scientific name.
Answer:
finding Cepheid variable and measuring their periods.
Explanation:
This method is called finding Cepheid variable and measuring their periods.
Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.
A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.
