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GenaCL600 [577]
4 years ago
5

In what direction relative to a magnetic field does a charged particle move in order to experience maximum deflecting force? Min

imum deflecting force?
Physics
1 answer:
grigory [225]4 years ago
4 0

Answer:

the particle experiences maximum force when it will be moving perpendicular to magnetic field.

it experiences minimum deflecting for when its going parallel to the magnetic field.

Explanation:when a charged particle is moving

magnetic field is B which is a vector

Magnetic force on a charge is F

formula for force  F =Q.(VxB)

magnitude of the charge particle is Q

the vector multiplication of (VxB) if most significant when the charge particle is moving parallel.Very minimum when moving

velocity of the charge particle is V

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Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c
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Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
E=k_e \frac{q}{r^2}
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C

While the electric field generated by the second charge is
E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C
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3 years ago
When a driver presses the brake pedal, his car can stop with an acceleration of 5.4 m/s2. How far will the car travel while comi
notsponge [240]
Given:\\a=5.4 \frac{m}{s^2}  \\v_0= 25\frac{m}{s} \\\\Find:\\s=?\\\\Solution:\\\\s=v_0t- \frac{at^2}{2} \\\\a= \frac{\Delta v}{t} \\\\v_e=0\Rightarrow \Delta v=v_0\\\\t= \frac{v_0}{a} \\\\s= \frac{v_0^2}{a} - \frac{v_0^2}{2a} =\frac{v_0^2}{2a} \\\\s= \frac{ (25\frac{m}{s})2 }{2\cdot 5.4 \frac{m}{s^2} } \approx 57.9m\\\\C\rightarrow Correct\;answer
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4 years ago
What average force is required to stop an 1100-kg car in 8.0s if the car is travelling at 95km/h?
Semenov [28]

Answer: 13062.5 N

What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h? i got the answer to be 13062.

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3 years ago
50 POINTS PLZ HELP<br> What does a plant use in photosynthesis and what are the products?
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A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of
4vir4ik [10]

Answer:

A) 2

Explanation:

Given;

magnetic field of the coil, B = 1.6 T

frequency of the coil, f = 75 Hz

maximum emf developed in the coil, E = 56.9 V

area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²

The maximum emf in the coil is given by;

E = NBAω

Where;

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N = E / BAω

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N = 2 turns

Therefore, the value of N is 2

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7 0
3 years ago
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