All categories

3 years ago

How do the distances between the orbits of the inner planets compare to the distances between the orbits of the outer planets?

1 answer:

3 0

Image result for How do the distances between the orbits of the inner planets compare to the distances between the orbits of the outer planets?

The inner planets are closer to the Sun and are smaller and rockier. ... The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars.

You might be interested in

Should i make my pet shark bark like a dog woof woof

mr_godi [17]

Answer: YES !!

Explanation: THAT WOULD BE AMAZING!!!

8 0

3 years ago

a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s

Mariulka [41]

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

= 6/15

So,

15x = 6x + 6d

9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

= (2/3) (d/dt)

Also,

Given:

d(d)=dt

= 7 ft/s

Thus,

d(d + x)=dt

= (7/3)d (d/dt)

Substitute, d= 7

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

4 0

3 years ago

30 POINTS!

brilliants [131]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

2 years ago

Explains this: a light source can emit more than one type of light

Radda [10]

Answer:

The electromagnetic spectrum comprise a lot of waves length. Usually, different waves length are called as different lights, and a light source can emit in more than a different wave length, as the sun does, for example. The sun emit the visible light, UV light, infrared, etc.

6 0

2 years ago