Judge the following items as true or false:

A particle of mass 10 g and charge 72 μC moves through a uniform magnetic field, in a region where the free-fall acceleration is

sineoko [7]

Answer:

$-0.07163\hat k\ T$ or 0.07163 T into the page

Explanation:

m = Mass of particle = 10 g

a = Acceleration due to gravity = -9.8j m/s²

v = Velocity of particle = 19i km/s

q = Charge of particle = 72 μC

B = Magnetic field

Here the magnetic and gravitational forces on the particle are applied in the opposite direction so,

$F_b=F_g$

$F_b=qvBsin\theta\\\Rightarrow F_b=qvBsin90\\\Rightarrow F_b=72\times 10^{-6}\times 19000B$

$F_g=ma\\\Rightarrow F_g=0.01\times -9.8$

$72\times 10^{-6}\times 19000B=0.01\times -9.8\\\Rightarrow B=\frac{0.01\times -9.8}{72\times 10^{-6}\times 19000}\\\Rightarrow B=-0.07163\hat k\ T$

The magnetic field is 0.07163 T into the page

5 0

2 years ago

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$

this expression holds for the points located at

-r₁ <x₁ <r₁

3) Point between the two spheres

E_ {total} = $k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}$

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

E_ {total} = $+ k \frac{Q}{(d-x_1)^2}$ + k Q / (d-x1) 2

point range

-r₂ <x₂ <r₂

5) Right side point

E_ {total} = $k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}$

E_ {total} = $- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )$ - k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0

2 years ago

Each blade of a fan has a radius of 11 inches. If the fan’s rate of turn is 1440o /sec, find the following. (a) The angular spee

notka56 [123]

Answer:

a) 24.43 radians per second

b) 268.73 inches per second

Explanation:

a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

$\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}$

$\omega = 1400\frac{C}{s}=24.43\frac{rad}{s}$

b) Linear speed on a point of the blade is related with angular speed of the fan by the equation

$v=\omega r$

with v linear speed, ω angular speed and r the radius of the blades. So:

$v=(24.43\frac{rad}{s})(11 in)$

Radians isn't really a unity; it is dimensionless so we can put it or not. So:

$v=268.73\frac{in}{s}$

3 0

3 years ago

Give an example of hypothesis for an experiment and then identify its dependent and independent variables. Write all the steps o

e-lub [12.9K]

An example of a hypothesis for an experiment might be: “A basketball will bounce higher if there is more air it”

Step one would be to make an observation... “hey, my b-ball doesn’t have much air in it, and it isn’t bouncing ver high”

Step two is to form your hypothesis: “A basketball will bounce higher if there is more air it”

Step three is to test your hypothesis: maybe you want to drop the ball from a certain height, deflate it by some amount and then drop it from that same height again, and record how high the ball bounced each time.

Here the independent variable is how much air is in the basketball (what you want to change) and the dependent variable is how high the b-ball will bounce (what will change as a result of the independent variable)

Step four is to record all of your results and step five is to analyze that data. Does your data support your hypothesis? Why or why not?

You should only test one variable at a time because it is easier to tell why the results are how they are; you only have one cause.

Hope this helps!

6 0

2 years ago

Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl

Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = ( $\frac{(m₂-m₁)}({m₁+m₂} * g$ = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

6 0

3 years ago