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svp [43]
3 years ago
5

What is Initial temperature and final temperature equations?? ...?

Physics
1 answer:
Neporo4naja [7]3 years ago
8 0
One that can help you is:
ΔT=<span>T<span>Final</span></span>−<span>T<span>Initia<span>l
That is of course adding both tmepratures. There is one more that is a lil bit more complex 
</span></span></span><span><span>Tf</span>=<span>Ti</span>−Δ<span>H<span>rxn</span></span>∗<span>n<span>rxn</span></span>/(<span>C<span>p,water</span></span>∗<span>m<span>water</span></span>)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps</span>
You might be interested in
The rules and expectations concerning correct or polite behavior is called _______________________.
aalyn [17]
The answer would be etiquette!
7 0
2 years ago
An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti
Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

  • There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
  • Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
  • So, we can write the following expression for Fg:

       F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)

b)

  • The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

       F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N  (2)

c)

  • According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

       a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2  (3)

3 0
2 years ago
Does kinetic energy stay the same at all heights? pls help I have 12 minutes to finish the project and the teacher won't help me
Alex Ar [27]

Answer:

Kinetic energy does not stay the same at all heights

Explanation:

Well as the height and wind increase so does the kinetic energy it's like when you fall as you are about to hit the floor you speed increases

HOPE THIS HELPS YA :)

7 0
3 years ago
A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each
nataly862011 [7]

The centripetal acceleration is 13.0 m/s^2

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference (2\pi r) and the period of revolution (T), so it can be rewritten as

v=\frac{2\pi r}{T}

Therefore we can rewrite the acceleration as

a=\frac{4\pi^2 r}{T^2}

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

T=\frac{1}{4}s = 0.25 s

Substituting into the equation, we find the acceleration:

a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0
3 years ago
The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by w
notsponge [240]

Explanation:

The given data is as follows.

         Resistance (R) = 1200 ohm,     Area (A) = 20 \times 10^{-6} m (as 1 \mu m = 10^{-6} m)

            Diameter (d) = 2.3 mm = 2.3 \times 10^{-3} m

First, we will calculate the length as follows.

            R = \rho \frac{L}{A}

Here,  \rho = resistivity of aluminium = 2.65 \times 10^{-8}

Putting the given values above and we will calculate the value of length as follows.

               R = \rho \frac{L}{A}

             1200 = 2.65 \times 10^{-8} \times \frac{L}{20 \times 10^{-6}}

               L = 9.056 \times 10^{5}

As the circumference of circular wire = 2 \pi r

or,                                                          = 2 \times \pi \times \frac{d}{2}  

                                                              = \pi \times d

And, number of turns will be calculated as follows.

             No. of turns × Circumference = Length of wire

              No. of turns × 3.14 \times 2.3 \times 10^{-3} = 9.056 \times 10^{5}

                               = 1.25 \times 10^{8}

Thus, we can conclude that 1.25 \times 10^{8}  turns of wire are needed.

6 0
3 years ago
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