The compound LiBr is an example of

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

C, N, Ne, Ar which contains a metal, nonmetal, noble gas , metalloid

postnew [5]

Answer:

c,carbono nonmetal

n, nitrogen nonmetal

ne, neón noble gas

ar,argon noble gas

4 0

2 years ago

The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t>0 , where is measured in seconds and

8090 [49]

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at $0$ .

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>

A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

$v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s$

The velocity is positive at when the time of motion is as follows;

$0$ .

The total distance traveled in the first 10 seconds is calculated as follows;

$2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m$

Learn more about motion of particles here: brainly.com/question/11066673

4 0

2 years ago

A double charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separa

Zanzabum

Answer:

$E=8*10^5\frac{V}{m}$

Explanation:

The magnitude of the electric field between two parallel conducting plates is defined as:

$E=\frac{\Delta V}{d}$

Here $\Delta V$ is the potential difference between the plates and d its separation.

The electric potential energy is defined as the product between the particle's charge and the potential difference:

$U=q\Delta V$

Solving for $\Delta V$ and replacing in the electric field formula:

$\Delta V=\frac{U}{q}\\E=\frac{U}{qd}$

In this case we have a double charged ion, so $q=2e$ :

$E=\frac{32*10^3eV}{(2e)(2*10^{-2}m)}\\E=8*10^5\frac{V}{m}$

6 0

2 years ago

Determine a valid way of finding the wire’s diameter if you know the resistivity of the material, \rho , and can measure the cur

olganol [36]

Answer:

To find the diameter of the wire, when the following are given:

Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.

Using the ohm's law,

Resistance R = (rho*L)/A

R = V/I.

Crossectional area of the wire A = π*square of radius

Radius = sqrt(A/π)

Diameter = Radius/2 = [sqrt(A/π)]

Making A the subject of the formular

A = (rho* L* I)V.

From the result of A, Diameter can be determined using

Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7

Explanation:

Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation

7 0

3 years ago