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3 years ago

3. What would the seasons be like if Earth's orbit had the eccentricity value of Mars?

2 answers:

4 0

According to Kepler's laws, the mass of a planet has almost no effect on its orbit; the mass of the sun is what controls things. Even though Earth is 10 times heavier than Mars, it would still trundle along Mars's old path. Both Mars and Earth are perpetually falling toward the sun, and all falling bodies fall at the same rate.

densk [106]3 years ago

3 0

Answer:

123456789

Explanation:

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What happens to light as it moves at an angle into a medium that has a lower index of refraction?

Aleks04 [339]

Answer:

It speeds up, and the angle increases.

Explanation:

6 0

4 years ago

A very long insulating cylinder of charge of radius 2.50 cm carries a uniform linear density of 15.0 nC/m. (

Pavlova-9 [17]

Answer:

r2 = 2.401557 cm

distance = 0.10 cm

Explanation:

given data

radius = 2.50 cm

density = 15.0 nC/m

voltmeter read = 175

solution

we know here potential difference that is express as

ΔV = $\frac{\lambda }{2\pi \epsilon _o} ln\frac{r2}{r1}$ ...........1

so here

$ln\frac{r2}{r1} = 2\pi \epsilon _o \times \frac{\triangle V}{\lambda }$

as here $\lambda$ is linear charge density

$\frac{r2}{r1} = e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}$

r2 = r1 × $e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}$

r2 = 2.40 × $e^{2\pi 8.85\times 10^{-12} \times \frac{175}{15\times 10^{-6} }}$

r2 = 2.401557 cm

and

here distance above surface will be

distance = r1 - r2

distance = 2.50 - 2.40

distance = 0.10 cm

4 0

3 years ago

Two identical metal balls a and b are mounted on insulating rods. Ball a has a charge of +q/2 and ball b is initially uncharged.

oksano4ka [1.4K]

Answer:

Help me please?

Explanation:

Did you get the answer? I believe it’s either C. +q or D. 0

6 0

3 years ago

Read 2 more answers

A current of 6.93 A 6.93 A in a long, straight wire produces a magnetic field of 4.15 μT 4.15 μT at a certain distance from the

natka813 [3]

Answer:

0.334 m

Explanation:

The magnetic field due to current in a straight thin wire, $B$ , is given by

$B = \dfrac{\mu_0 I}{2\pi R}$

where $\mu_0$ is the permeability of free spave with value $4\pi\times10^{-7}$ and $R$ is the distance from the wire.

$R = \dfrac{\mu_0 I}{2\pi B}$

Substituting the values from the question,

$R = \dfrac{4\pi\times10^{-7}\times 6.93}{2\pi\times 4.15\times 10^{-6}}$

$R = \dfrac{2\times10^{-1}\times6.93}{4.15} = 0.334 \text{ m}$

7 0

4 years ago

Read 2 more answers

a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t

lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0

3 years ago