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tekilochka [14]
1 year ago
8

In physics class, Anne is designing a small circuit that can operate a mixer. Identify the devices she should use in the circuit

.
Physics
1 answer:
Lubov Fominskaja [6]1 year ago
7 0

To operate a mixer circuit , Anne should use a (motor, buzzer, bulb, or battery) to transfer electric energy into motion. She should also use a (switch, resistor, buzzer, or bulb) to start and stop the flow of current.

An electric-powered circuit consists of a device that gives energy to the charged particles constituting the modern-day, consisting of a battery or a generator; devices that use cutting-edge, including lamps, electric cars, or computer systems; and the connecting wires or transmission traces.

An electrical circuit is an interconnection of electrical components or a version of such an interconnection, consisting of electrical elements. an electrical circuit is a network along with a closed loop, giving a go-back direction for the cutting-edge.

An electrical circuit consists of a source of electrical strength, two wires that can convey electric power present day, and a mild bulb. One stop of both the wires is hooked up to the terminal of a mobile at the same time as their free ends are related to the light bulb. the electric circuit is damaged while the bulb is switched off.

Learn more about the circuit here:-brainly.com/question/2969220

#SPJ1

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Whats a good air time and maximum height
serg [7]

Answer: Time to reach max. height is the time taken by it to reach the topmost position whereas time of flight is the overall time taken by it, i.e. till it reach down back to the thrower or launcher. And also time of flight is double the time to reach maximum height.

Explanation:

5 0
3 years ago
Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5
AlladinOne [14]

Answer:

a)432000\frac{N}{C}\hat{i}

b)-6.92\times10^{-14}N\hat{i}

Explanation:

a)

The magnitude of the electric field generated by a charged particle at a distance r is:

E= k\frac{|Q|}{r^{2}}

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is 25.0cm-5.0cm=20.0 cm=0.2m apart the positive charge is:

E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}

E_1= 45000\frac{N}{C}

and the electric field generated by q2:

E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}

E_2=216000\frac{N}{C}

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}

\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}

Vector quantities satisfy superposition principle, this is \overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}, with E the total electric field.

\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}

b) The force is:

\overrightarrow{F}=e*\overrightarrow{E},

with q the charge of an electron

\overrightarrow{F}=(-1.61\times10^{-19})*(432000)\hat{i}=-6.92\times10^{-14}N\hat{i}

8 0
3 years ago
Do you think distance and time are relevant terms in describing motion?
olya-2409 [2.1K]

Answer:

yes

Explanation:

because motion is relevant

6 0
3 years ago
Read 2 more answers
Energy generated from tides is called ____________________.
aalyn [17]
Energy generated from tides is called tidal power, or tidal energy, and it is a form of hydropower that transforms energy obtained from tides into electricity, mainly. 
8 0
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The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t
jek_recluse [69]

Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

4 0
4 years ago
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