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3 years ago

Moira is drawing the electric field lines around a pair of charges. One charge is positive, and the other charge is negative.

Physics

2 answers:

kati45 [8]3 years ago

7 0

Answer:

Explanation:

i just took the test

GalinKa [24]3 years ago

5 0

The answer to the question, "How should she draw the field lines?" is option C, <span>away from the positive charge and toward the negative charge</span>

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A naked, male corpse was found at 8AM on Tuesday, July 9.The air temperature wasalready 26.7˚C (81˚F). The body exhibited some s

Slav-nsk [51]

Answer:

x = 9.87 hours ago.

Further explanation is present below.

Explanation:

Data given:

Air Temperature = 26.7°C (81°F)

Body temperature = 34.4°C (93.9°F)

We are given that, Livor mortis was absent. Secondly, The body exhibited some stiffness in the face and eyelids.

So, it can be concluded that, body must have died recently, as Livor mortis was absent plus the presence of stiffness in the face and eyelids. Furthermore,

The normal body temperature = 37°C

Body temperature in our case = 34.4° C

Difference = 37 -34.4 = 2.6°C

And Air temperature = 26.7°C

Which is a very low difference, in addition, it has not acquired the air temperature so, all this evidence certifies that body died recently.

Now, we have to calculate, how recently it died.

Formula:

0.78 . x = difference of body temperature and air temperature

Where, x = number of hours = time required,

0.78 . x = (34.4 - 26.7)

0.78 . x = 7.7

x = 7.7/0.78

x = 9.87 hours ago.

It means that, naked male corpse was died 9.87 hours ago.

Now, we are asked to find out clothing on the body had made a difference or not.

So, Yes it would definitely make a difference. As body would have more warmer than being naked. and it would result in a calculation of a different time of death. (body temperature would have > 34.4°C)

3 0

3 years ago

During a softball game, a shortstop catch a ground ball. The action forces is the ball pushing on the glove. What is the reactio

marusya05 [52]

<span>The reaction force would be the exact opposite of the action. In this case, choice (c) would be the most correct. If the action is the ball pushing the glove, the reaction would then be the glove pushing back on the ball.</span>

4 0

3 years ago

I need some advice I have two girlfriends as of now.

Paul [167]

Answer:

I Would go with Kye.

Explanation:

Why i would go with her is because she has more life experience with you. She also knows you better. I would usually go with the closest one and the one who knows you the best.

Hope this helps.

<3 Have a good day!

4 0

3 years ago

Read 2 more answers

Whats a transverse wave?

Mazyrski [523]

Answer:

a wave vibrating at right angles to the direction of its propagation.

Explanation:

7 0

4 years ago

An AC source of maximum voltage V0 = 30 V is connected to a resistor R = 50 Ω, an inductor L = 0.6 H, and a capacitor C = 20 µF.

ivolga24 [154]

Hello!

We can begin by solving for the resonance ANGULAR frequency of the circuit.

For an RCL circuit, the resonance angular frequency is given as:
$\omega_0^2 = \frac{1}{LC}\\\\w_0 = \sqrt{\frac{1}{LC}}$

ω₀ = resonance angular frequency (rad/s)

L = Inductance (0.6 H)
C = Capacitance (20 μF)

Plug in the values and solve.

$\omega_0 = \sqrt{\frac{1}{(0.6)(0.00002)} } = 288.675 \frac{rad}{s}$

For an AC power source, the output is usually expressed as:

$V(t) = V_{max}sin(\omega_0 t})$

So, using the appropriate values and setting the source angular frequency equivalent to the circuit's resonance angular frequency:

$V(t) = 30sin(288.675t)$

To find the maximum charge on the capacitor when the frequency of the source is equivalent to the resonance frequency of the circuit (or the angular frequencies are equal), we can begin by finding the maximum voltage across the capacitor.

To find this, however, we must solve for the maximum current across the circuit by finding the total impedance of the circuit. When the circuit is at resonance, the impedance is equivalent to the resistance of the RESISTOR.

So, solve for the maximum current in the circuit using Ohm's Law:

$i = \frac{V}{R}$

In this instance AT RESONANCE:

$I_{Max} = \frac{V_Max}{R}\\\\I_{Max} = \frac{30}{50} = 0.6 A$

Now, we must solve for the capacitive reactance in order to find the maximum voltage across the capacitor. Using the following equation for capacitive reactance:
$X_c = \frac{1}{\omega C}\\\\X_c = \frac{1}{(288.675)(0.00002)} = 173.205 \Omega$

Now that we found the maximum current and capacitive reactance, we can now solve for the maximum voltage across the capacitor:
$V_{C, max} = X_C I_{Max}\\\\V_{C, max} = 173.205 * 0.6 = 103.923 V$

Finally, we can easily solve for the maximum charge on the capacitor using the relationship:
$C = \frac{Q}{V}\\\\Q = CV$

Plug in the values solved for above.

$Q = (0.00002)(103.923) = 0.00208 C = \boxed{2.078 mC}$

6 0

2 years ago