¿De qué altura cae un cuerpo que tarda 4 s en llegar al suelo? usar g = 10 m/s2 *

A wheelbarrow is a good example of a second-class lever. True or False

olganol [36]

Answer:

true

Explanation:

a wheelbarrow has its load situated between the fulcrum and the force the wheel Barrow is 2nd class because of its resistance between the force and the axis

3 0

2 years ago

Based on the diagram which of the following statements is true of a helium atom

snow_lady [41]

Answer: Option 2

Explanation:

Option 1 is wrong because there are 2 protons and 2 neutrons in nucleus.

Option 3 is wrong because there are two electrons moving around nucleus.

Option 4 is wrong because electrons are negatively charged and are moving around the nucleus.

Option 5 is wrong because there equal amount of protons and electrons with 2 each.

8 0

2 years ago

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)