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3 years ago

Pleaswe help me ewwfea

Physics

2 answers:

Talja [164]3 years ago

6 0

Answer:

The first one is Earths Tilt on its axis

amm18123 years ago

5 0

1. B
I’m not sure about the second one ): sorry

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In rules for naming a compound you change the negative name to end in" ide" *

Strike441 [17]

Answer:

Single-atom negative ions end in “-ide”, so binary compounds always have this ending. Polyatomic compounds usually end in “-ate” or “-ite”. FORMULAS: Write the positive ion, with its charge, then the negative ion, with its charge.

Explanation:

hope it helps, please mark as brainliest

8 0

2 years ago

) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in

Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

$T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}$

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0

3 years ago

A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.

raketka [301]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, $a=4\ m/s^2$

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(30)^2}{2\times 4}$

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

6 0

2 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

2 years ago

How do the chemical properties of the halogens compare to those of the noble gases?

serg [7]

Halogens<span> are extremely reactive elements because they need one more electron to gain a full octet of valence electrons, whereas the </span>noble gases<span>are extremely unstable because they already have their full octet.</span>

8 0

3 years ago