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suter [353]
3 years ago
11

You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10

−19 C. Answer in units of µC.
Physics
1 answer:
AleksandrR [38]3 years ago
6 0

Answer:

1.92×10^-9 microCoulomb

Explanation:

Elemental charge = 1.6×10^-19 Coulomb

Charge of balloon after 12,000 electrons have been removed from it = 12,000 × 1.6×10^-19 Coulomb = 1.92×10^-15 Coulomb = 1.92×10^-15/10^-6 = 1.92×10^-9 microCoulomb

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A rocket sled with an initial mass of 3 metric tons, invluding 1 ton of fuel, rests on a level section of track. At t=0, the sol
ELEN [110]

Answer:

v = 719.2 m / s and     a = 83.33 m / s²

Explanation:

This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is

           v - v₀ = v_{e} ln (M₀ / M)

where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket

In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s

            m_fuel = 75 10

            m_fuel = 750 kg

As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is

              M = 3000 -750 = 2250 kg

let's calculate

            v - 0 = 2500 ln (3000/2250)

            v = 719.2 m / s

To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is

             Push = v_{e} dM / dt

let's calculate

             Push = 2500  75

             Push = 187500 N

            If we use Newton's second law

             F = m a

             a = F / m

let's calculate

              a = 187500/2250

              a = 83.33 m / s²

7 0
3 years ago
What speed should a satellite of mass 4,900 kg moving around
Rudiy27

Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.

<u>Given the following data:</u>

  • Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
  • Mass of Moon = 7.36 × 10²² kg
  • Distance, r = 4.2 × 10⁶ m.

<h3>How to determine the speed of this satellite?</h3>

In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.

This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:

Fc = Fg

mv²/r = GmM/r²

<u>Where:</u>

  • m is the mass of the satellite.
  • M is mass of the Moon.

Making v the subject of formula, we have;

v = √(GM/r)

Substituting the given parameters into the formula, we have;

v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)

v = √(1,168,838.095)

v = 1,081.13 m/s.

Speed, v = 1.8 × 10³ m/s.

Read more on speed here: brainly.com/question/20162935

#SPJ1

8 0
2 years ago
sports photographers often use large aperture, long focal length lenses. what limitations do these lenses impose on the photogra
Brut [27]

Answer:

The depth of focus achievable with those lenses is small.

Explanation:

A larger aperture makes it much harder to focus on more than one object. When using a telephoto lens (the ones the question is referring to), the depth of focus is very small. For example, using a telephoto lens to take a photo of a runner might get the runner in focus, but certainly not the track, or the audience behind them. If you look at photos, especially older photos, of Olympians in almost any sport you can see this.

Hope this helps!

6 0
3 years ago
How do scientists classify species?
serious [3.7K]

Answer:

D. By comparing traits

Explanation:

Because age isn't genetic, as well as names, as well as who discovered, but traits are genetic.  

6 0
3 years ago
Read 2 more answers
Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
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