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BigorU [14]
3 years ago
11

Juan compró un carro que dicen que es muy rápido. Cuándo lo probó recorrió una distancia de 4500m en tan solo 5 min. ¿Qué veloci

dad tiene el carro?
Physics
1 answer:
sweet [91]3 years ago
4 0

Answer:

Don't know sorry..................

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec
solmaris [256]

To solve this problem it is necessary to apply the concepts related to Wavelength depending on the frequency and speed of light.

The equation that meets these parameters is given by

\lambda = \frac{c}{f}

Where,

c = Speed of light

\lambda = Wavelength

f = Frequency

Our values are given as

c = 3*10^8m/s \rightarrow velocity of light in free space

f = 2Hz

Replacing we have that

\lambda = \frac{c}{f}\\\lambda = \frac{3*10^8}{2}\\\lambda = 1.5*10^8m

Therefore the correct answer is B.

3 0
3 years ago
If an object has the same number of positive and negative changes, it’s electrical charge is ?
Marrrta [24]

The electrical charge is neutral. I hope this helps!

8 0
3 years ago
Which organism in the food web above is sometimes a first-level consumer and sometimes a second-level consumer?Explain
scoray [572]
The answer is: Mouse/Herbivore.

First organism is always a producer (plant) such as grass.
The 2nd organism is the first level or primary consumer. Ex. Mouse - It eats the producer, so it is a herbivore. 
4 0
3 years ago
Once a falling object has reached a constant velocity, the object ___.
Rashid [163]

'a',  'b',  and  'c'  are all reasonable statements.

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Read 2 more answers
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
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