Physics <br> I’ll give brainiest! :)

Explain what causes spring and neap tides, and identify the phases of the moon when these tides occur. Be sure to mention what h

slega [8]

Answer: Spring tides occur when the moon is full or new. Earth, the moon, and the Sun are in a line. The moon’s gravity and the Sun’s gravity pull Earth’s crust and ocean water. This causes tides to be higher than normal.

At neap tide, the moon and the Sun are at right angles to each other. This happens during the first and third quarters of the lunar cycle. At neap tide, the Sun’s gravity and the moon’s gravity are balanced. High tides are lower; low tides are higher.

Explanation:

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7 0

3 years ago

Read 2 more answers

Which symbol and unit of measurement are used for electric current?

Burka [1]

Answer: Symbol is I and unit A

Explanation: A represents Amperes

HOPE THIS HELPS!!!!!!!!

4 0

3 years ago

Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12

Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

$F_C = k\frac{C_1C_2}{R^2}$

where k = $8.99\times10^9 nm^2/C^2$ is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

$120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}$

$C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}$

Since their total charge is 200C:

$C_1 + C_2 = 200$ or $C_1 = 200 - C_2$

We can substitute the above equation

$C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}$

$-C_2^2 +200C_2 - 2.13\times10^{-9} = 0$

$C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}$

$C= \frac{-200\pm200}{-2}$

$C = 1.06 \times 10^{-11}$ or $C \approx 200$

So the larger charge is C = 200 C

8 0

4 years ago

You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a vo

S_A_V [24]

Answer:

A. Z = 185.87Ω

B. I = 0.16A

C. V = 1mV

D. VL = 68.8V

E. Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

$Z=\sqrt{R^2+\omega^2L^2}$ (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

$Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega$

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

$I=\frac{V}{Z}$ (2)

V: voltage amplitude = 30.0V

$I=\frac{30.0V}{185.87\Omega}=0.16A$

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

$V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV$ (3)

D. The voltage across the inductor is:

$V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V$

E. The phase difference is given by:

$\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°$

5 0

3 years ago

A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

AVprozaik [17]

To solve this problem it is necessary to take into account the concepts related to the magnetic moment and the torque applied over magnetic moments.

For the case of the magnetic moment of a loop we have to,

$\mu = IA$

Where

I = Current

A = Area of the loop

Moreover the torque exerted by the magnetic field is defined as,

$\tau = IAB$

Where,

I = Current

A = Area of the loop

B = Magnetic Field

PART A) First we need to find the perimeter, then

$P = 2\pi r$

$r = \frac{P}{2\pi}$

$r = \frac{1.9}{2\pi}$

$r = 0.3025m,$

The total Area of the loop would be given as,

$A = \pi r^2$

$A = \pi 0.3025^2$

$A = 0.287m^2$

Substituting at the equation of magnetic moment we have

$\mu = (16*10^{-3})(0.287)$

$\mu = 4.58*10^{-3} A.m^2$

Therefore the magnetic moment of the loop is $4.58*10^{-3}Am^2$

PART B) Replacing our values at the equation of torque we have that

$\tau = IAB$

$\tau = (16*10^{-3})(0.287)(0.790)$

$\tau = 3.62*10^{-3}Nm$

Therefore the torque exerted by the magnetic field is $3.62*10^{-3}Nm$

6 0

3 years ago

Physics I’ll give brainiest! :)