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2 years ago

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To properly operate the laboratory thermometer Group of answer choices it needs to be shaken but make sure you have enough room

to shake it safely there is no need to shake it before measurement always shake it before measurement

1 answer:

ruslelena [56]2 years ago

5 0

Answer:

it needs to be shaken but make sure you have enough room to shake it safely

Explanation:

To properly operate the laboratory thermometer it needs to be shaken but make sure you have enough room to shake it safely. This done because there is a small bend in the mercury channel of a clinical thermometer that uses mercury. You must shake the thermometer to get the mercury from a previous reading from the thermometer back into the bulb for taking new reading. The bend prevents flow back into the tube so that one can comfortably take reading.

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Mineral salt compounds (calcium chloride, table salt)

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Answer:

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Explanation:

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An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p

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Answer:

a) $v=1.77\times 10^6\ m/s$

b) $v=3.872\times 10^6\ m/s$

c) $v=5.5\times 10^6\ m/s$

d) $v=6.7\times 10^6\ m/s$

e) $v=7.7\times 10^6\ m/s$

Explanation:

Given that

d = 2 cm

V = 200 V

$u=4\times 10^5\ m/s$

We know that

F = E q

F = m a

E = V/d

So

m a = q .V/d b

$a=\dfrac{q.V}{m.d}$ ---------1

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=1.6\times 10^{-19}\ C$

Now by putting the all values in equation 1

$a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2$

$a=1.5\times 10^{15}\ m/s^2$

We know that

$v^2=u^2+2as$

a)

s = 0.1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}$

$v=\sqrt{3.16\times 10^{12}}\ m/s$

$v=1.77\times 10^6\ m/s$

b)

s = 0.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}$

$v=\sqrt{1.5\times 10^{13}}\ m/s$

$v=3.872\times 10^6\ m/s$

c)

s = 1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}$

$v=\sqrt{3.06\times 10^{13}}\ m/s$

$v=5.5\times 10^6\ m/s$

d)

s = 1.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}$

$v=\sqrt{4.5\times 10^{13}}\ m/s$

$v=6.7\times 10^6\ m/s$

e)

s = 2 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}$

$v=\sqrt{6.06\times 10^{13}}\ m/s$

$v=7.7\times 10^6\ m/s$

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Answer:

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Explanation:

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The force of gravity is (-8 N) and the the normal force is 8 N.

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Here there is force of 30 N is applied in horizontal direction. The frictional force act in opposite direction. So, the force of friction is -30 N so that box across the floor.

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Answer:

A.

Explanation:

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