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1 year ago

A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?

Physics

1 answer:

Ilya [14]1 year ago

8 0

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

= 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

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A transformer is used to light a lamp rated 40w, 240v from a 400v A.C supply. Calculate:

zubka84 [21]

Answer:1.81

(a) Explanation:the turn ratio= input voltage÷output voltage.

400÷220=1.81.

Don't know how to solve b part...

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3 years ago

A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located

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Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

$v = \lambda f$

here we know that the wavelength of the wave is

$\lambda = 2.5 m$

$f = 10 Hz$

now speed of the wave is given as

$v = 10(2.5)$

$v = 25 m/s$

Now time taken by the wave to reach 5 m distance is

$t = \frac{L}{v}$

$t = \frac{5}{25}$

$t = 0.2 s$

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3 years ago

What are scientists who study space called?

Brilliant_brown [7]

Answer:

Astronomer

Explanation:

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3 years ago

Read 2 more answers

To balance a chemical equation we use______. -coefficients -subscripts

Vitek1552 [10]

Answer:

Coefficients

Explanation:

Coefficients are used to balance chemical equations.
According to the law of conservation of mass, the mass of the reactants in a chemical equation should be equivalent to the mass of the products.
The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products.

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3 years ago

A very long conducting cylinder (length L) of radius R(R<R and (b)r

mina [271]

Answer:

Explanation:

We have to find electric potential V at a distance r.

a) For r>R,

The electric field in the cylinder is given by

E.A equating it to the other electric field given by

б.A/ε₀

Here the area of cylinder is given by= 2*3.14*r*L

While for the outside, the area= 2*3.14*R*L

Equating both, we get

E= бR/rε₀

Now,

The potential difference is given as:

ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.

Where ΔV= V₀-V

So solving we get

V₀=V-бR/ε₀ln (r/R)

b) For r<R i.e. inside the cylinder

There will be no electric field produced as E=0

So ultimately Vin= V

c) V=0 at r= infinity.

4 0

3 years ago