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professor190 [17]
4 years ago
8

Marta, Cato, and Juan plan a skit to illustrate one of the methods of charging. Marta and Cato are walking together. Each has on

e flower and one box of candy. Juan offers Cato a huge bouquet of flowers, but Cato can’t take it because his hands are full. Cato hands his flower to Marta and then turns to Juan to take the bouquet. Marta walks away in one direction with two flowers and one box of candy. Juan keeps the bouquet and walks away in the other direction. Cato looks sad standing alone with only a box of candy.
How does the skit model a method of charging?

A) It models induction because an electron (a flower) transfers from Cato to Marta through direct contact.

B) It models friction because an electron (a flower) transfers from Cato to Marta through direct contact.

C)It models friction because the offer of flowers represents a charged object that causes an electron (a flower) to transfer from one object to another.

D) It models induction because the offer of flowers represents a charged object that causes an electron (a flower) to transfer from one object to another.
Physics
1 answer:
sveta [45]4 years ago
8 0

Answer:

Option D.

Explanation:

Methods of charging:

1. By friction: when two materials of different electro-negativities rub against each, electrons transfer from one material to another.

2. By induction: When a charged body is brought near an uncharged body, the later gets polarized.

3. By conduction: when electrons from one body to another uncharged while contact, it is known as charging by conduction.

Here, a flower is analogous to an electron. A box of candy is positive charge. A bouquet of flowers is like a charged body which is brought near an uncharged body. It induces charge on ti it by causing it to transfer electron to another.

Thus, option D is correct.

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A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantit
Bess [88]

Answer:

C. 85%

Explanation:

A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?

A. 15%

B. 30%

C. 85%

D. 100%

work done by the system will be

W=PdV

p=pressure

dV=change in volume

3tam will be changed to N/m^2

3*1.01*10^5

W=3.03*10^5*(1.5-1)

convert 0.5L to m^3

5*10^-4

W=3.03*10^5*5*10^-4

W=152J

therefore

to find the percentage used

152/1000*100

15%

100%-15%

85% uf the fuel's energy was lost to friction and heat

6 0
3 years ago
How many stars are there in an Open Cluster?*
Step2247 [10]

Answer:

Very few

Most open clusters form with at least 100 stars

Brainly, please.

4 0
3 years ago
At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
3 years ago
Sound waves that enter the ear canal are directed to the ____, causing it to vibrate.
andre [41]

According to the research, the correct option the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.

<h3>What is the eardrum?</h3>

It is the membrane found in the middle ear of vertebrate animals, separating this sector from the external auditory canal.

When sound waves enter through the external auditory canal, the eardrum vibrates, transmitting its movement to the middle ear through a series of bones and in this way, the pressure change is transformed into a mechanical movement.

Therefore, we can conclude that according to the research, the correct option is the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.

Learn more about the eardrum here: brainly.com/question/12770491

#SPJ1

4 0
2 years ago
Question #5: Explain how a reflecting telescope is different from a refracting telescope. List the two different types of reflec
Anni [7]

The main component in a reflecting telescope is a mirror where the light will bounce off and is then focused into a smaller area. In contrast, a refracting telescope uses lenses that focus the light as it travels towards the other end.

Two different types of reflecting telescopes are:

1.Cassegrain reflector

2.Newtonian telescope

Explanation:

  • The distinction between the two is in how they manipulate the incoming light in order to magnify the image. The main component in a reflecting telescope is a mirror where the light will bounce off and is then focused into a smaller area.
  • Key advantage of reflecting telescopes is how big you can make them. With lenses, the maximum size is limited to about one meter, largely because of the problems stated above as well as the skyrocketing costs.
  • The Newtonian telescope, also called the Newtonian reflector, is a type of reflecting telescope invented Sir Isaac Newton, using a concave primary mirror and a flat diagonal secondary mirror. The Newtonian telescope's simple design has made it very popular with amateur telescope makers.
  • The Cassegrain reflector is a combination of a primary concave mirror and a secondary convex mirror, often used in optical telescopes and radio antennas, the main characteristic being that the optical path folds back onto itself, relative to the optical system's primary mirror entrance aperture.

6 0
3 years ago
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