At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972

4 0

3 years ago

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

Marina86 [1]

Answer: $\alpha =10.66 rad/s^2$

Explanation:

Given

mass of disk $m=5 kg$

diameter of disc $d=30 cm$

Force applied $F=4 N$

Now this force will Produce a torque of magnitude

$T=F\cdot r$

$T=4\dot 0.15$

$T=0.6 N-m$

And Torque is given Product of moment of inertia and angular acceleration $(\alpha )$

$T=I\cdot \alpha$

Moment of inertia for Disc $I= \frac{Mr^2}{2}$

$I=0.05625 kg-m^2$

$0.6=0.05625\cdot \alpha$

$\alpha =10.66 rad/s^2$

3 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)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago