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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

An object of mass 100kg is raised 2m above the ground using an Inclined Plane of length 10m calculate the effort parallel to inc

MArishka [77]

Answer:

Slope = 2 m / 10 m = 1/5

For every 5 m of effort the object will be raised 1 m

W = work done on object = M g h increase in PE of object

E S = W where E is effort and S the distance thru which the effort acts

E S = M g H

E = 100 kg * 9.8 m/s^2 * 2 m / 10 m = 196 kg m / s^2 = 196 N

Check: total work = 2 * 9.8 * 100 = 1960 J

Force Needed = 1960 J / 2 m = 980 Newtons

Mechanical advantage = 980 / 196 = 5 as one would expect since the object is raised 1 m for every 5 m of force input

8 0

2 years ago

Question 5 of 10

Nina [5.8K]

It is D 14 MS. West. Kauehfbfnd

6 0

3 years ago

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

Alexandra [31]

Answer:

The maximum no. of electrons- $2.25\times 10^{22}$

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

$I = \frac{Q}{t}$

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

$Q = 1.0\times 3600 = 3600\ C$

Maximum number of electrons, n is given by:

$n = \frac{Q}{e}$

where

e = charge on an electron = $1.6\times 10^{- 19}\ C$

Thus

$n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}$

3 0

3 years ago

Physics question!!!

lesantik [10]

Answer:

4.5 metres

Explanation:

Using Hooke's Law ( $F\propto x$ )

We need to find the spring constant of the bungee cord with the given extension and force, we can do this by substituting in known values.

$600=3k\\k=200$

Now we have found the spring constant of the bungee cord, we can substitute it in for the a different force. As the cord is the same we can use the same spring constant.

$900=200x\\x=4.5$

5 0

3 years ago