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A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= $2pie\frac sqrt {m}{k}$

$\frac {{f2}/times {fo}}$ =1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒ $\frac{f2} {f1}$ = $\frac sqrt{m1}[m2}$

⇒ $\frac{f2}{fo}$ = 1:3

⇒f2= $\frac{fo} {3}$

6 0

3 years ago

If carbon has an atomic number of 6, how many protons and neutrons are found in the carbon-14 atom? A.

Alex Ar [27]

The correct answer is B. 6 protons and 8 neutrons

Carbon-14 has same atomic number of 6. It has a nucleon number of 14
Atomic number = proton number = 6
Neutron number = nucleon number - atomic number = 14 - 6 = 8

Hope it helped!

5 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)