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amm1812
3 years ago
8

Which shows the formula for converting from kelvins to degrees Celsius? °C = (9/5 × K) +32 °C = 5/9 × (K – 32) °C = K – 273 °C =

K + 273
Physics
2 answers:
madam [21]3 years ago
5 0

The freezing point of the water is 0 C , and it equals to 273 K

Then, To convert from Kelvins degrees to Celsius degrees we use the relation

K = C + 273

Also,

C = K - 273

mafiozo [28]3 years ago
4 0
<h2>Answer with Explanation </h2>

The temperature T in degrees Celsius (°C) is equal to the temperature T in

Kelvin (K) minus 273.15.

Formula: °C =K - 273.15

Degrees Celsius ‘°C’ and kelvins ‘K’ have the same magnitude. Only the starting points are different in these two scales. In the Kelvin scale, 0 K is absolute zero, whereas in the Celsius scale, 0°C is the freezing point of water. So, by definition, it means absolute zero (0 K) is equivalent to −273.15 °C.

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What type of material is wrapped around an object could cause the object to lose the most heat
Dmitrij [34]
<span>A sheet of copper could cause the object to lose the most amount of heat. Copper is an essential element and a good conductor of heat. Heat can transfer from one end of a piece of copper to the other end.</span>
7 0
3 years ago
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
3 years ago
A train runs from New Delhi to Hyderabad it covers first of 420 km in 7 hours and the next distance of 360 km in 6 hours​
Mkey [24]

Explanation:

the answer is 16.7 miles or 60kmph

5 0
3 years ago
Read 2 more answers
Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)
Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No

N = 0.1 No

Let the time taken is t and the decay constant is λ.

The relation between the decay constant and the half life is given by

\lambda =\frac{0.6931}{T}=\frac{0.6931}{8}=0.08664 per day

Us the equation of radioactivity

N=N_{0}e^{-\lambda t}

0.1N_{0}=N_{0}e^{-0.08664 t}

e^{0.08664 t}=10

Taking natural log on both the sides, we get

0.08664 t = 2.303

t = 26.6 days

4 0
3 years ago
Which element most likely interacts with water the same way lithium interacts with water?
oksano4ka [1.4K]

Answer:

Is there a multiple choice or select all that apply? I would say Potassium (K) or Sodium (Na)

Explanation:

6 0
2 years ago
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