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Assoli18 [71]
2 years ago
6

A 6.80 $\mu C$ particle moves through a region of space where an electric field of magnitude 1230 N/C points in the positive $x$

direction, and a magnetic field of magnitude 1.32 T points in the positive $z$ direction. If the net force acting on the particle is 6.18E-3 N in the positive $x$ direction, calculate the magnitude of the particle's velocity. Assume the particle's velocity is in the $x$-$y$ plane.
Physics
1 answer:
Murljashka [212]2 years ago
5 0

Answer:

v = -227.785 m/s

Explanation:

The electric field exerts the following force on the electric particle:

F = qE

F = 6.65 \times 10^{-6} \times 1230

F = 0.0081795 \ N

The magnetic field exerts the following force on the particle::

F = qvB

F = 6.65\times 10^{-6} \times v \times 1.32

F = 8.778 \times 10^{-6} \times v

Total force acting is:

F = qvB + qE

6.18 \times 10^{-3} = 0.0081795 + 8.778 \times 10^{-6} \times v

v = \dfrac{6.18 \times 10^{-3}  -0.0081795 }{8.778 \times 10^{-6}}

v = -227.785 m/s

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A charged particle moves through a magnetic field. In which situation is the magnetic force zero?
maksim [4K]

Answer:

The answer is the option a.

Explanation:

We know that magnetic force (Fm) is defined as

Fm = q (v x B)

Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.

"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is

v x B = vBsin (thetha)

So,

Fm = qvBsin (thetha)

And, in case in which v and B are parallel vectors, thetha is zero, and,

sin (thetha)=sin (0) = 0

So, Fm=0

7 0
3 years ago
Write the percent as a fraction or a mixed number in simplest form. <br><br><br> 7%
svetlana [45]
The actual answer is 7/100
4 0
3 years ago
A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,
Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a)  Kinetic energy of block = potential energy in spring  

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg.  Spring constant k is unknown, but you can find it from given data:  

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.  

From the energy equation above, solve for v,

v = v √(k/m)  

= 0.15 √(300/1)

= 2.598 m/s.

b)  Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.  

This is the same momentum carried by bullet as it strikes the block.  Therefore, if u is bullet speed,  

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0
3 years ago
A car traveling on a straight road at 15.0 meters per second accelerates uniformly to a speed of 21.0 meters per second in 12.0
oksano4ka [1.4K]
<h2>Option 3,  216 m is the correct answer.</h2>

Explanation:

We have initial velocity, u = 15 m/s

Time, t = 12 seconds

Final velocity, v = 21 m/s

We have equation of motion v = u + at

Substituting

                     21 = 15 + a x 12

                       a = 0.5 m/s²

Now we have equation of motion v² = u² + 2as

                           21² = 15² + 2 x 0.5 x s

                            s = 216 m

       Displacement = 216 m

Option 3,  216 m is the correct answer.

8 0
3 years ago
Bob is moving at 0.967c with respect to Alice. At the exact instant he passes Alice, she fires a very short laser pulse in the s
yulyashka [42]

Explanation:

Speed of Bob, v = 0.967 c

At the exact instant he passes Alice, she fires a very short laser pulse in the same direction Bob is moving.

(a) We need to find the distance measured by Alice  between Bob and the laser pulse. It is given by :

d=ct-vt

d=t(c-v)

d=5.59\times (c-0.967c)

d=5.53\times 10^7\ meters

(b) Distance measured by Bob between himself and the laser pulse is given by :

d_B=ct

d_B=3\times 10^8\times 5.59

d_B=6.67\times 10^9\ meters

Hence, this is the required solution.  

8 0
3 years ago
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