In which situation does the energy constantly change from gravitational potential energy to kinetic energy?

Two balls of equal mass at the bottom of an incline are rolled upward without slipping at the same initial velocity. one ball is

german

The appropriate response would be the hollow ball.The purpose for it has more kinetic energy at the last (2/3 as opposed to 2/5) so it will have more potential energy (more height) at the best. Pondering it, more inertia implies its less impervious to change meaning the hollow ball with more inertia wont need to stop as fast.

3 0

4 years ago

A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

6 0

3 years ago

X-rays with an energy of 86 keV scatter off of stationary electrons. How much energy do the photons scattering at 60° have? A. 7

AveGali [126]

Answer:

A. 79.3 keV

Explanation:

Because the procedure involves many steps for its resolution and it works faster on paper and pencil, the detailed solution of this exercise is attached as a scanned image of the procedure for review.

In the procedure, the initial values of the problem and the replacement of these values with the correct formulas for this process are taken into account.