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oksian1 [2.3K]
4 years ago
14

Hurryyyyyy When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand p

oints in the direction that the charge is moving? palm
fingers
back
thumb
Physics
2 answers:
masya89 [10]4 years ago
6 0
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.

</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
Ronch [10]4 years ago
3 0

Answer: The fingers.

Explanation: The magnetic force on a charge is given by the equation:

F = q(VxB)

where q is the charge of the particle, V is the velocity of the particle and B is the magnetic field.

If you use the right-hand method, you need to point with the tip of your fingers to the direction of the velocity, face the palm of your hand in the direction of the magnetic field, and now your thumb will point in the direction of the magnetic force.

So the part of the hand that points in the direction that the charge is moving is the second option, "fingers"

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A sailboat is traveling across the ocean with a speed of 4 m/s when large gusts of wind pick up that starts to accelerate the sa
Ivahew [28]

solution:

Using Cartesian co-ordinate system

Final velocity =v= -4m/s

Initial velocity = u

Acceleration a = m/s²

Time (t).= 60s

By the first kinematical equation

V= u +at

U = v – at

=(-4)-(-3)(60)

176m/s

So, initial velocity was 176m/s


8 0
4 years ago
How does the mass of an object change the gravitational pull on two objects in a system? How does the distance between objects i
LuckyWell [14K]

Answer:The greater the mass, there greater then gravitational pull on their objects.

There greater the distance,the lower the gravitational pull between the object

Explanation:

Gravitational pull(force) is directly proportional to the products Of The masses. therefore if the mass increase,the gravitational pull(force) also increases.

Gravitational pull(force) is inversely proportional to distance.if the distance between the objects increases,the gravitational pull(force) decreases .

4 0
3 years ago
Give v=120m/s north and t = 10.2, calculate the displacement
Sedbober [7]

Answer:

Displacement from the starting position is 1224m

Explanation:

The object is moving at 120 meters per second, it does this for 10.2 seconds. Use multiplication to find the answer.

120 \times 10.2 = 1224

4 0
3 years ago
PLEASE HELLPPP ITS FIR A FINAL 3.Speed (v) abs wavelength ( λ ) and the frequency (f) of sound are related asRequired to answer.
Aliun [14]

Answer:

B. v = λ X f

<em><u>Hope this helps! </u></em>

8 0
3 years ago
A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0
4 years ago
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