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3 years ago

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What is another name for Kepler's first law? A. law of periods B. law of ellipses C. law of universal gravitation D. law of equa

l areas in equal times

1 answer:

pashok25 [27]3 years ago

8 0

B. Law of ellipses. which states that the path of planets about the Sun is in Elliptical shape (hence the name).

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How long does it take for 4 coulombs of charge to pass through a cross

Galina-37 [17]

Answer: 2 seconds

Explanation:

Given that,

Time (T) = ?

Charge (Q) = 4 coulombs

current (I) = 2 Amps

Since charge depends on the amount of current flowing through the wire in a given time, hence

Charge = Current x Time

Q = IT

4 coulombs = 2 Amps x Time

Time = 4 coulombs / 2 Amps

Time = 2 seconds

Thus, it takes 2 seconds for the current to flow through the wire

4 0

3 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i

a_sh-v [17]

Answer:

a)<em> It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves. </em>

b) <em>The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds. </em>

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

Gamma rays

X-rays

Ultraviolet rays

Visible region

Infrared

Microwave

Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of $3x10^{8}m/s$ .

<em>a) Find the time it took for his voice to reach the Earth via radio waves.</em>

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

$c = \frac{d}{t}$ (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

The distance from the Earth to the Moon is $3.85x10^{8} m$ , therefore.

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

<em>b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.</em>

The distance from the Earth to the Mars at its closest approach is $5.76x10^{10}m$ , therefore.

$t = \frac{5.76x10^{10}m}{3x10^{8}m/s}$

$t = 192s$

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0

3 years ago

On the kelvin scale what is the freezing point of water

Alla [95]

Freezing point of the water is known as 273 K

Hope this helps!

8 0

3 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$

Replacing,

$a =\frac{42N}{83000kg}$

$a =5.06*10^{-4}m/s^2$

The total speed change

$\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity }$ we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

$a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}$

We have that,

$t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }$

$t = 1403.16s$

Therefore the Rocket should be fired around to 1403.16s

7 0

3 years ago

What the density of an object that has a volume of 20ml and mass of 3g.

Oxana [17]

Answer:

The equation of D = m/V

Where D = density

m = mass

and V = volume

We are solving for V, so with the manipulation of variables we multiply V on both sides giving us

V(D) = m

now we divide D on both sides giving us

V = m/D

We know our mass which is 600g and our density is 3.00 g/cm^3

so

V = 600g/3.00g/cm^3 = 200cm^3 or 200mL

a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL

If you wish to change it to L, you'd have to convert

Explanation:

8 0

3 years ago