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3 years ago

Which of the answer choices best defines mixture?

Physics

1 answer:

Svetlanka [38]3 years ago

3 0

Combination of two or more substances that can be separated using only chemical means

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4. A high school athlete runs 400 m in 52.20 sec. What is his speed in mph? 14100 1110

Serggg [28]

Answer:

28,800m/p second

Explanation:

Calculate the distance per second so, 400m/50 s= 8m/p second now knowing the speed/hour and knowing an hour has 3,600 seconds,multiply it by 8 then you will get 28,800m/p second, or 28.8km/h

3 0

3 years ago

Nobel gases share which characteristics?

Leto [7]

Full outer shell of electrons, they are colorless and odorless ,their melting and boiling points are close together which gives them very narrow liquid range

8 0

3 years ago

Read 2 more answers

A code of ethics for psychologists helps to do all of the following except __________.

s2008m [1.1K]

I'm pretty sure the answer is B: establish public doubt
Hope this helps!

6 0

3 years ago

Read 2 more answers

Which is a characteristic of an amorphous solid?

Degger [83]

Answer:

Becomes softer as the temperature rises!

Explanation:

4 0

3 years ago

Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106V/m ) if

xxMikexx [17]

(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

(b) The plates can be close together to 1.7 mm with this applied voltage

Explanation:

Given data:

Dielectric strength of air = $3 \times 10^{6} V / m$

Distance between the plates = 2.00 mm = $2.00 \times 10^{-3} \mathrm{m}$

Potential difference, V = $5.0 \times 10^{3} V$

We need to find

a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not

b) the minimum distance at which the plates can be close together with this applied voltage.

The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,

$E=\frac{V}{d}=\frac{5.0 \times 10^{3}}{2.00 \times 10^{-3}}=2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$

From the above, concluding that The electric field strength between two parallel conducting plates ( $2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$ ) does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

b) To find how close together can the plates be with this applied voltage:

The formula would be,

$d_{\min }=\frac{V}{E_{\max }}$

Apply all known values, we get

$d_{\min }=\frac{5.0 \times 10^{3}}{3 \times 10^{6}}=1.7 \times 10^{-3} \mathrm{m}=1.7 \mathrm{mm}$

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3 years ago