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levacccp [35]
3 years ago
8

At what water temperature will additional heat energy need to be added before the temperature will change again?

Physics
1 answer:
Katyanochek1 [597]3 years ago
7 0
That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)

Hope this helps! (If correct, please rank as brainliest answer) :)
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True or false : Scientific theories have withstood the test of time and are accepted as proven fact.
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Gravitational potential energy is related to–A. what an object is made of. B.an object’s motion.C. an object’s position.D. an ob
Bas_tet [7]

Answer:

The Answer is B because the material the object is made of, the position, or the color have absolutely nothing to do with gravitational potential energy

7 0
3 years ago
A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie
ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, A=7.4\ cm^2=7.4\times 10^{-4}\ m^2

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, B_1=0.5\ T

Final magnetic field, B_2=3\ T

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_2-B_1}{t}

\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}

\epsilon=1.65\times 10^{-3}\ V

Induced current in the loop of wire is given by :

I=\dfrac{\epsilon}{R}

I=\dfrac{1.65\times 10^{-3}}{2.4}

I=6.87\times 10^{-4}\ A

So, the induced current in the loop of wire over this time is 6.87\times 10^{-4}\ A. Hence, this is the required solution.

7 0
3 years ago
Consider the standing wave pattern below created by a string fixed at both ends. The string is under a tension of 0.98 N and has
Shalnov [3]

Answer:

11.07Hz

Explanation:

Check the attachment for diagram of the standing wave in question.

Formula for calculating the fundamental frequency Fo in strings  is V/2L where;

V is the velocity of the wave in string

L is the length of the string which is expressed as a function of its wavelength.

The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)

Therefore L = 1.5λ

If L = 3.0m

1.5λ = 3.0m

λ = 3/1.5

λ = 2m

Also;

V = √T/m where;

T is the tension = 0.98N

m is the mass per unit length = 2.0g = 0.002kg

V = √0.98/0.002

V = √490

V = 22.14m/s

Fo = V/2L (for string)

Fo = 22.14/2(3)

Fo = 22.14/6

Fo = 3.69Hz

Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo

Frequency of the wave = 3×3.69

Frequency of the wave = 11.07Hz

3 0
4 years ago
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