Explanation:
It is given that,
Mass of an object is 13 kg
We need to find the mass of this object on the surface of Jupiter.
The mass of an object is the amount present in it. It it independent of location. It mass remains the same at Jupiter as well. Its does not depend on the gravity on that surface.
Hence, its mass will remain 13 kg on the surface of Jupiter.
I know the answer,I just learned this,U measure it by hertz
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
First, let's calculate the frequency of this radiation, which is given by:
![f= \frac{c}{\lambda}](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7Bc%7D%7B%5Clambda%7D%20)
where c is the speed of light and
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
is the photon wavelength. For this radiation, photons have wavelength of
![\lambda=13.0 cm=0.13 m](https://tex.z-dn.net/?f=%5Clambda%3D13.0%20cm%3D0.13%20m)
Therefore their frequency is
![f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{0.13 m}=2.3\cdot 10^9 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7Bc%7D%7B%5Clambda%7D%3D%20%5Cfrac%7B3%20%5Ccdot%2010%5E8%20m%2Fs%7D%7B0.13%20m%7D%3D2.3%5Ccdot%2010%5E9%20Hz%20%20)
The energy of a photon with frequency f is given by
![E=hf](https://tex.z-dn.net/?f=E%3Dhf)
where h is the Planck constant. By using the frequency we found before, we find the energy of a single photon of this radiation: