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Aneli [31]
3 years ago
11

Two positively charged particles are separated on an axis by 0.1 meters, and q1 has a 6 µC charge and q2 has a 2 µC charge. Reca

ll that k = 8.99 × 109 N • meters squared over Coulombs squared.. What is the force applied between q1 and q2 in N? In which direction does particle q2 want to go?
Physics
2 answers:
ZanzabumX [31]3 years ago
8 0

Answer:

10.8 N & away from particle q1, in a straight line to the right

Explanation:

•°•°•°

Vlada [557]3 years ago
5 0

Answer:

Force between two charges is 10.8 N

both charges are positive nature charge so they both repel each other and hence q2 will move away from charge q1

Explanation:

As we know that the force between two charge particles is given by Coulomb's law of electrostatics

So we will say it as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = 6\mu C

q_2 = 2\mu C

d = 0.1 m

F = \frac{8.99 \times 10^9(6 \mu C)(2\mu C)}{0.1^2}

F = 10.8 N

Since both charges are positive nature charge so they both repel each other and hence q2 will move away from charge q1

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3 years ago
The periodic table lists all the elements according to the number of __________ they have! You can’t change __________ without c
raketka [301]

Answer:

protons

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5 0
2 years ago
A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie
Citrus2011 [14]

Answer:

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(b) 0.389 T

Solution:

As per the question:

Hall emf, V_{Hall} = 20\ mV = 0.02\ V

Magnetic Field, B = 0.10 T

Hall emf, V'_{Hall} = 69\ mV = 0.069\ V

Now,

Drift velocity, v_{d} = \frac{V_{Hall}}{B}

v_{d} = \frac{0.02}{0.10} = 0.2\ m/s

Now, the expression for the electric field is given by:

E_{Hall} = Bv_{d}sin\theta                            (1)

And

E_{Hall} = V_{Hall}d

Thus eqn (1) becomes

V_{Hall}d = dBv_{d}sin\theta

where

d = distance

B = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

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(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

5 0
2 years ago
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PilotLPTM [1.2K]

Answer:

10. 36 g ZnCl2

Explanation:

Zn + 2HCl  -> ZnCl2 + H2

0.076 mol Zn

1.37 mol HCl

3 mol H2

Limiting reactant: Zn

1 mol Zn        -> 1 mol ZnCl2

0.076 mol Zn  ->x                         x= 0.076 mol ZnCl2=10.36 g

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USPshnik [31]

Answer:

Explanation:

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