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3 years ago

Using the rules of significant rules what do you get when you add 24.545 and 307.3?

1 answer:

4 0

When adding and subtracting, significant figures rules say to have as many decimal places as the least amount in your data. So when you add them, round to one decimal place. This is because 307.3 only has one number after the decimal place.

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When can a high speed velocity cause damage?'

sweet-ann [11.9K]

Answer:

50 Mph.

Explanation:

According to the National Severe Storms Laboratory, winds can really begin to cause damage when they reach <em><u>50 mph</u></em>. But here’s what happens before and after they reach that threshold, according to the Beaufort Wind Scale (showing estimated wind speeds): - at 19 to 24 mph, smaller trees begin to sway.

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3 years ago

An atom has an atomic number of 12 and a mass amber of 25. How many neutrons does the stom have? (1 point)

Alisiya [41]

Answer:

23423421342

Explanation:

342342342

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3 years ago

What question could you ask about kinetic energy which will include the variables that affect it?

Juli2301 [7.4K]

KE= (1/2) mv²

So, the variables we need to include in our question would be a varable for a mass(m) of an object at some velocity(v).

My Answer:

(This is just an example question, yours can be different)
What is the Kinetic Energy experienced by an bouncy ball rolling at 7m/s (that's your velocity) across a frictionless surface that has a mass(m) of 10 grams?

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3 years ago

Read 2 more answers

At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

tino4ka555 [31]

Answer:

$W=\frac{p_0V_0-p_1V_1}{\gamma-1}$

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

$pV^{\gamma}=k$

where,

$p$ = pressure

$V$ = volume

$\gamma=\frac{C_p}{C_V}$

$k$ = constant

Therefore, work done by the gas during expansion is,

$W=\int\limits^{V_1}_{V_0} {p} \, dV$

$=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV$

$=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\$

(using $pV^{\gamma}=k$ )

$=\frac{p_0V_0-p_1V_1}{\gamma-1}$

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3 years ago

True or False: If a point charge has electric field lines that point into it, the charge must be positive.

Licemer1 [7]

Answer:

False

Explanation:

If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges

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3 years ago