The period-luminosity relation is critical in finding distances with

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

Zolol [24]

Considering the unknown resistence as R and using the Ohm's First Law, we have:

$i= \frac{V}{R_{eq}} \\ 0.5= \frac{12}{R+10} \\ R+10=24 \\ R=14-Ohm$

The equivalent resistence is given by the resistor series with the lamp resistence.

$R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}$

If you notice any mistake in my english, please let me know, because i am not native.

6 0

3 years ago

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horiz

igomit [66]

Answer:

The direction is due south

Explanation:

From the question we are told that

The energy of the electron is $E = 19.0keV = 19.0 *10^3 eV$

The earths magnetic field is $B = 42.3 \muT = 42.3 *10^{-6} T$

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

$\= F = q (\= v * \=B)$

On the first uploaded image is an illustration of the movement of the electron

Looking at the diagram we can see that in terms of direction the magnetic force is

$\= F =q(\=v * \= B)= -( -\r i * - \r k)$

$= -(- (\r i * \r k))$

generally i cross k = -j

so the equation above becomes

$\= F = -(-(- \r j))$

$= - \r j$

This show that the direction is towards the south

3 0

3 years ago

A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. No

kozerog [31]

Answer: $-2502N$

Explanation:

$(V_2)^2=(V_1)^2+2ad$

where;

$V_2$ = final velocity = 0

$V_1$ = initial velocity = 60 km/h = 16.67 m/s

$a$ = acceleration

$d$ = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

$60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s$