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emmainna [20.7K]
1 year ago
11

Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw

o 4.0-cm-diameter disks.
Physics
1 answer:
aalyn [17]1 year ago
4 0
Ok, I think this is right but I am not sure:
 Q = ϵ
 0AE
A= π π
 r^2

=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons


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Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father
-Dominant- [34]
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
6 0
2 years ago
Read 2 more answers
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
2 years ago
Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache
FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

      W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis  

      N - W cos θ  = 0

The equation for the force of friction is

      fr = μ N

      fr = μ mg cos θ

We replace at work

     W = - μ m g cos θ  d

Mechanical energy in the lower part of the embankment

      Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

     Em_{f} = U = m g y

     W = ΔEm =  Em_{f} - Em₀

    - μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

     sin θ = y / d

     y = d sin θ

   

    - μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

     d (g syn θ + μ g cos θ) = ½ v²

     d = v²/2 g (sintea + myy cos tee)

     d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

     d = 1555.85 /7.8145

     d = 199.1 m

Let's use trigonometry to find the height

      sin 16 = y / d

      y = d sin 16

      y = 199.1 sin 16

      y = 54.9 m

8 0
2 years ago
Which of the following statements cannot be supported by Kepler's laws of planetary motion?
gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio \frac{a^3}{T^2} (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

8 0
2 years ago
An activity that is relatively short in time (&lt; 10 seconds) and has few repetitions predominately uses the _____________ ener
tensa zangetsu [6.8K]
An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level. 
Steps of the ATP-PC system:


1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage.  This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.


2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.


3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP.  This newly made ATP can now be broken down to issue energy to fuel activity. 
5 0
2 years ago
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