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Alex Ar [27]
2 years ago
11

In Physics lab, Ellen has been given the task of constructing a simple motor. She must find common household items at home and b

ring them to class the following day. Her list of items include:
9 volt battery and leads, copper wire rotor, platform to hold the rotor, a magnet, and a wood base

She sets up her motor in physics lab the next day and finds that it works well. Which statement below explains the flow of energy through Ellen's motor?
A) battery-->electrical current-->copper wire rotor -->magnet--> mechanical energy
B) battery--> magnet--> electrical current--> copper wire rotor--> mechanical energy
C) magnet--> mechanical energy--> copper wire rotor--> battery--> electrical current
D) magnet--> copper wire rotor--> mechanical energy--> battery--> electrical current
Physics
2 answers:
Aleks [24]2 years ago
6 0

Answer:

a) battery-->electrical current-->copper wire rotor -->magnet--> mechanical energy

Explanation:

kiruha [24]2 years ago
4 0

Answer:

The answer is A.

Explanation:

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The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?
Klio2033 [76]
The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:
n= \frac{c}{v}
where the speed of light in vacuum is c=3 \cdot 10^8 m/s. The speed of light in benzene is v=2.00 \cdot 10^8 m/s, so we can use the previous relationship to find the refractive index of benzene:
n= \frac{3 \cdot 10^8 m/s}{2.00 \cdot 10^8 m/s}=1.5
7 0
2 years ago
On a cold winter day when the temperature is −20∘C, what amount of heat is needed to warm to body temperature (37 ∘C) the 0.50 L
vlabodo [156]

Answer:

75.6J

Explanation:

Hi!

To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,

Given the above we have the following equation.

Q=(m)(h2)-(m)(h1)

where

m=mass=1.3×10−3kg.

h2= entalpy at 37C

h1= entalpy at -20C

Q=m(h2-h1)

remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference

Q=mCp(T2-T1)

Cp= specific heat of air = 1020 J/kg⋅K

Q=(1.3×10−3)(1020)(37-(-20))=75.6J

4 0
2 years ago
Walt ran 5 kilometers in 25 minutes, going eastward. What was his average velocity?
storchak [24]
1km per 5 mins
PLEASE VERIFY WITH SOMEONE I MAY BE INCORRECT
7 0
2 years ago
Read 2 more answers
In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min
Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, t_{K} = 17.9 min = 17.9\times 60 = 1074 s

Time taken by Hannah, t_{H} = 15.3 min = 15.3\times 60 = 918 s

Now, the speed of Kara and Hannah can be calculated respectively as:

v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s

v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s

Time taken in each lap is given by:

(v_{H} - v_{K})t = x

(5.45 - 4.65)\times t = 400

t = \frac{400}{0.8}

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

d_{H} = v_{H}\times t

d_{H} = 5.45\times 500 = 2725 m

No. of laps taken by Hannah when she passes Kara:

n_{H} = \frac{d_{H}}{x}

n_{H} = \frac{2725}{400} = 6.8 ≈ 7 laps

3 0
2 years ago
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through
Svet_ta [14]

Answer:

The speed is \sqrt{2}v_{0}.

(a) is correct option.

Explanation:

Given that,

Potential difference V= V_{0}

Speed v = v_{o}

If it were accelerated instead

Potential difference V'=2V_{0}

We need to calculate the speed

Using formula of initial work done on proton

W = q V

We know that,

\Delta W=\Delta K.E

q V=\dfrac{1}{2}mv^2

Put the value into the formula

q V_{0}=\dfrac{1}{2}mv_{0}^2

v_{0}^2=\dfrac{2qV_{0}}{m}....(I)

If it were accelerated instead through a potential difference of 2 V_{0}, then it would gain a speed will be given as :

Using an above formula,

v_{0}'^2=\dfrac{2qV_{0}}{m}

Put the value of V_{0}

v_{0}'^2=\dfrac{2q\times2V_{0}}{m}

v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}

v_{0}'=\sqrt{2}v_{0}

Hence, The speed is \sqrt{2}v_{0}.

6 0
2 years ago
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