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2 years ago

In which direction does the electric field point at a position directly east of a positive charge

Physics

2 answers:

lora16 [44]2 years ago

7 0

Field lines always point away from the positive side of a magnet. So i would say east but im not to sure

natima [27]2 years ago

5 0

Answer: Towards East

Explanation:

The direction of the electric force per unit charge i.e. the electric field is given by the direction of motion of positive test charge under the electric force. The field lines are directed radially outwards for a positive charge and radially inwards for a negative charge.

Thus, the electric field points towards East for the position directly east of a positive charge

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Bill has a mass of 85 kg and is skating west. He increases his speed from 3 m/s to 5 m/s by applying a force for 3 seconds. What

likoan [24]

F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N

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2 years ago

A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the fi

allochka39001 [22]

Answer:

a. q2 = 16.4μC, positive charge

b. F = 0.900N

c. downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

$F_e=k\frac{q_1q_2}{r^2}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

$q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C$

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

3 0

2 years ago

A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting

Georgia [21]

Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

$\Delta U = n C_v(T_f-T_i)$

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

$\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K$

So, $\Delta U= 0.2\times12.5(1200-300)\\=2250 J$

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

$\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J$

Since, Q is positive 2250 J of heat will flow into the system.

6 0

2 years ago

If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v

dexar [7]

The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units

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2 years ago

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

bazaltina [42]

Look on this website http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html

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2 years ago