Honors — June 5 to 6 homework

A 500-turn rectangular loop of wire has an area per turn of 4.5 × 10-3 m2. At t0 = 0 s, a magnetic field is turned on, and its magnitude increases to 0.50 T when t = 0.75 s. The field is directed at an angle of φ= 30.0o with respect to the normal of the loop. (a) Find the magnitude of the average emf induced in the loop. (b) If the loop is a closed circuit whose resistance is 6.0 Ω, determine the average induced current.

AP — Optional Assignment

There are a lot of units and terminology associated with radiation, as shown on this site. My question is, can you explain what they mean? This is an optional, one-page assignment with a mandatory first draft before the final hand-in. It is due by Monday June 9th by the end of school (the first draft) emailed to my school email, mvannucci@frhsd.com.

Please type it in Word!

Homework for night of May 28

To My AP:

Read from page 939 to 949 and take notes on it. This is the entirety of the relativity we will be doing, so we will do a quick review and some problems Monday and then a short test Tuesday.

Tomorrow, May 29th you all will be in room D104 prepping for your symposia. This is a last-minute change but the goal is to help you prepare.

To my Honors:

The force between two wire lengths can be derived from the equation you learned in class AND:

F=I_2 ∆LB where I_2 is the second current.

AP — Weekend HW

Orbital Speed Equation

Consider a satellite with mass Msat orbiting a central body with a mass of mass MCentral. The central body could be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object. If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship

Fnet = ( Msat • v2 ) / R

This net centripetal force is the result of the gravitational force that attracts the satellite towards the central body and can be represented as

Fgrav = ( G • Msat • MCentral ) / R2

Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force can be set equal to each other. Thus,

(Msat • v2) / R = (G • Msat • MCentral ) / R2

Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by dividing through by Msat. Then both sides of the equation can be multiplied by R, leaving the following equation.

v2 = (G • MCentral ) / R

Taking the square root of each side, leaves the following equation for the velocity of a satellite moving about a central body in circular motion

where G is 6.673 x 10-11 N•m2/kg2, Mcentral is the mass of the central body about which the satellite orbits, and R is the radius of orbit for the satellite.

(source, The Physics Classroom)


Given the above, what would the orbital velocity of a person at sea level be? What about at Low Earth Orbit? (These are altitudes you’ll have to look up). At what orbit will a satellite’s orbital velocity match the rotation speed of Earth, or about 460 m/s?