All categories

3 years ago

What is the difference in the positioning of the Sun, Earth and moon for a lunar eclipse versus a full moon?

1 answer:

3 0

When the Moon passes between Sun and Earth, the lunar shadow is seen as a solar eclipse on Earth. When Earth passes directly between Sun and Moon, its shadow creates a lunar eclipse. Lunar eclipses can only happen when the Moon is opposite the Sun in the sky, a monthly occurrence we know as a full Moon.

You might be interested in

A girl runs once around a circular track with a radius of 100m at speed of 10m/s.

goblinko [34]

Answer:

0 m

Explanation:

Displacement is the shortest distance from one ppont to another. Here, the girl returns to the same spot after running. Hence, displacement is 0 m.

5 0

3 years ago

Si un cuerpo adquiere una carga de -0,02 C, ha ganado o ha perdido electrones? Cuantos?

sammy [17]

Question :

If a body acquires a charge of -0.02 C, has it gained or lost electrons? Many?

Solution :

We know, charge gained is shown by negative sign.

Since, charged acquired is given as -0.02 C .

Therefore, it is body has gained electrons.

Now, number of electrons is given by :

$n = \dfrac{net\ charge}{charge \ on \ one \ electron}\\\\n = \dfrac{-0.02}{-1.60 \times 10^{-19}}\\\\n = 1.25\times 10^{17}$

Hence, this is the required solution.

7 0

3 years ago

Mercury, a planet in our Solar System, and Callisto, a moon of the planet Jupiter, are about the same size. However, Mercury’s m

TiliK225 [7]

You may jump higher because the more the mass of the planet, the more gravitational force. There is less mass(and gravity) on Callisto so you wouldn’t be weighed down as much and can jump higher. Whereas on Jupiter there is more weight holding you down.

3 0

3 years ago

Read 2 more answers

An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.

bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

p = 1.7 10⁻⁹ 6.8 10⁻⁶

p = 1.16 10⁻¹⁴ C m

The potential energie dipole is described by the expression

U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

θ = 0º

U = 1.16 10⁻¹⁴ 1160 cos 0

U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

θ = 180º

cos 180 = -1

U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

ΔU = | U1 -U2 |

ΔU = 1.35 10-11 - (-1.35 10-11)

ΔU = 2.7 10 -11 J

6 0

3 years ago

An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =

Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V $_m$ = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³ )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V $_m$ = V₀sin( ωt )

we substitute

25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

$I$ (t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

$I$ (t) = V₀√(C/L)

we substitute

$I$ (t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

$I$ (t) = 25 × 0.00002

$I$ (t) = 0.0005 A

$I$ (t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0

3 years ago