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3 years ago

Describe 2 characteristics of the neap tide. *

Physics

1 answer:

Mariana [72]3 years ago

5 0

Answer:

Neap tides, which also occur twice a month, happen when the sun and moon are at right angles to each other. ... This occurs twice each month. The moon appears new (dark) when it is directly between the Earth and the sun. The moon appears full when the Earth is between the moon and the sun.

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I'm trying to find the potential energy of a planet using formula G M(sun) x m(planet) / r. I have given G - newton's graviation

ludmilkaskok [199]

Answer:

$-5.39\times10^{33} J$

Explanation:

Potential energy =

$-\frac{GMm}{r}\\=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times5.97\times10^{24}}{147/1\times10^{9}}\\=-5.39\times10^{33} J$

7 0

2 years ago

A parallel-plate capacitor is constructed from two 6.0 cm × 6.0 cm electrodes spaced 1.5 mmapart. The capacitor plates are charg

Brilliant_brown [7]

A) $2.4\cdot 10^{-6} J$

The energy stored in a capacitor is given by:

$E=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored

C is the capacitance

The capacitance of a parallel-plate capacitor is

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=6.0 cm \cdot 6.0 cm=36.0 cm^2=36\cdot 10^{-4} m^2$ is the area of each plate

$d=1.5 mm=0.0015 m$ is the distance between the plates

Substituting,

$C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0015 m}=2.1\cdot 10^{-11} F$

The charge stored on the capacitor is

$Q=10 nC=10\cdot 10^{-9}C$

So, the energy stored is

$E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{2.1\cdot 10^{-11} F}=2.4\cdot 10^{-6}J$

B) $2.6\cdot 10^{-6}J$

This time, the separation between the plates is

d = 1.7 mm = 0.0017 m

So, the new capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0017 m}=1.9\cdot 10^{-11} F$

And so, the new energy stored is

$E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{1.9\cdot 10^{-11} F}=2.6\cdot 10^{-6}J$

Energy must be conserved, so the difference between the initial energy of the capacitor and its final energy is just equal to the work done to increase the separation between the two plates from 1.5 mm to 1.7 mm (in fact, the two plates of the capacitor attract each other since they have opposite charge, so work must be done in order to increase their separation)

7 0

3 years ago

You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe

steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, $L=800\ mH=800\times 10^{-3}\ H=0.8\ H$

Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

$Z=\sqrt{R^2+(X_C-X_L)^2}$ ...........(1)

Where

$X_C$ is the capacitive reactance, $X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega$

$X_L$ is the inductive reactance, $X_L={2\pi fL}$

$X_L={2\pi \times 60\times 0.8}=301.59\ \Omega$

So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

Z = 106.26 ohms

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

8 0

4 years ago

List the types of radiation in the electromagnetic spectrum from lowest to highest

Ahat [919]

Radio waves, Television waves, Microwaves, Infrared light, Visible light, Ultraviolet rays, X - rays, Gamma rays, and i think thats about it.

Sorry if i get this wrong Im in elementry school
Hope this helped :)

8 0

4 years ago

Please help!!! I will mark brainliest asap, this is due in 10 minutes!

VARVARA [1.3K]

Answer:

speed

(distance per time)

Kathy is faster btw

4 0

3 years ago

Read 2 more answers