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2 years ago

What is the value of the fixed points of a temperature scale

Physics

1 answer:

AleksAgata [21]2 years ago

8 0

Answer:

To calibrate a thermometer is to mark a thermometer so that you can use it to measure temperature accurately. A fixed point is a standard degree of hotness or coldness such as the melting point of ice or boiling point of water.

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A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat, parallel plates of glass (n = 1.50). What mu

Harrizon [31]

Answer:

t = 96.1 nm

Explanation:

For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength

now we know that the path difference of two reflected light from thin liquid layer is given as

$2\mu t - \frac{\lambda}{2} = N\lambda$

here we know that

$\mu = 1.756$

t = thickness of layer

N = 0 (for minimum thickness of layer)

$\lambda = 675 nm$

now we have

$2(1.756) t = \frac{675 nm}{2}$

$t = 96.1 nm$

5 0

2 years ago

Segmented mirrors sag under their own weight. their optical shape must be controlled by computer-driven thrusters under the mirr

madam [21]

Active Optics.

Hope that helps, Good luck! (:

6 0

2 years ago

What type of clouds are associated with low pressure?

Naddik [55]

Cumulus and cumulonimbus<span />

5 0

3 years ago

Solve for x. 4(x+1)≤4x+3

alexdok [17]

4x + 4 < 4x + 3 (expand it)
4 < 3 (cancel 4x on both sides)
Since 4 < 3 is not true there is no solution.

Answer: NO SOLUTION.

3 0

3 years ago

Read 2 more answers

. A ball is thrown downward at a speed of 20 m/s. Choosing

a_sh-v [17]

The final velocity is $v = 20 - gt$

The distance traveled by the ball at time t is $y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is $0 = (20)^2 - 2g(y-y_0)$

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

$v = 20 - gt$

The distance traveled by the ball at time t;

$y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is calculated as;

$v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)$

Learn more here: brainly.com/question/16878713

3 0

2 years ago