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allsm [11]
3 years ago
13

Jeff’s father is installing a do-it-yourself security system at his house. He needs to get a device from his workshop that conve

rts electric energy to sound energy. Which device is Jeff’s father looking for? A. switch B. motor C. buzzer D. bulb E. battery
Physics
2 answers:
elena55 [62]3 years ago
7 0

The answer is a buzzer. hope this helps. this is the answer because it use electricity to make the sound please award brainliest


luda_lava [24]3 years ago
5 0

Answer:

C). Buzzer

Explanation:

In electric buzzer is a device which produce sound of high amplitude when electric current flows through it.

As per the concept of energy we know that the electrical energy when given to the buzzer then the electrical energy is converted into sound energy.

So here we can say that the correct answer here is

C).Buzzer

while in other options

B) Motor : Electrical energy is converted into mechanical energy

D) Bulb : electrical energy is converted into light energy

E) Battery : Chemical energy is converted into electrical energy

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In a basketball game, a player shoots a jump shot. Which force actually causes the player to jump? the player pushing down on th
Gemiola [76]

Answer:  the player pushing down on the floor

Explanation:

4 0
3 years ago
Read 2 more answers
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
1. How do galaxies change over time? 2. How are galaxies created? 3. How was Hubble important to our understanding of galaxies?
VashaNatasha [74]

Hi!


1. Galaxies are constantly evolving with time, and one manner that they do this is that they change their rotation. This is apparent from the knowledge we have of initially formed galaxies which are spiral like in the beginning, but over time transform into what seems like smooth, barren disks. Another difference is that some galaxies are younger than others, and the age difference is observed with younger galaxies having a higher number of stars that are also comparatively brighter, which makes these galaxies look blue. Older galaxies contain older stars, and older stars are seen to give off red light, which is why these galaxies seem red. Galaxies also undergo various physical and chemical changes, and older galaxies are more structured.


2. Galaxies are brought into formation with the major factor being gravity. The force of gravity is responsible for attracting various stars, collapsing gas, dust clouds and other dark matter into a structured body. Over time the organization becomes more defined. So basically, a galaxy starts off as clouds of dust and stars in space, which come into proximity with other similar clouds, and their subsequent interactions shape the structure of the galaxy.


3. Edwin Hubble, also known as the 'pioneer of distant stars', has contributed to out understanding of galaxies in that he was the first person to prove the existence of galaxies other than the Milky Way, through his studies on spiral nebulae (which were formerly thought to be clouds of dust and gas, but were actually galaxies). He also established the relationship of red shift of a galaxy (recession velocity) and the distance of the galaxy from the Earth - a directly proportional relationship, which is known as the Hubble's Law.


4. The Milky Way is found to be considerably larger, about ten times, than the average dwarf galaxy being 100,000 light years across in diameter. The Milky Way is home to 200 billion stars, and with the capacity (dust and gas) to give rise to billions more. The Milky Way is the biggest galaxy in the Virgo Supercluster, and is relatively old in comparison with most other galaxies within the universe.


5. Dark matter and gravity play a role in determining the structure of the galaxy and influences the environment over time. Dark matter contributes to the mass of the galaxy, and this affects the spin of the galaxy. The more the mass of a galaxy (or more massive a galaxy) the greater would be its rotation. Gravity is the acting force that holds the components of galaxy in place, such as the dust clouds, colonies of star, gas and dark matter. Gravity influences the structure in that it exerts and inward pull on the stars, resulting in the spiral arm structure of a galaxy, which over time tends to be more eclipse like.


Hope this helps!

7 0
3 years ago
you read a primary source and a secondary source that discuss the same experiment. There is a difference in the conclusion made
Nuetrik [128]
You should trust the primary source more.

This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.

The primary source should be trusted more because it is from direct observation.     
4 0
3 years ago
Which is a valid velocity reading for an object?
pashok25 [27]

Answer:

45 m / s North is a valid vector reading for an object.

Explanation:

Then velocity will be defined by x km / hr North. And, magnitude of velocity defines the speed of the body. Although this tells the speed, but there is no description for the direction, so it's not a vector reading

5 0
2 years ago
Read 2 more answers
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