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MAXImum [283]
3 years ago
9

During a camping trip, Sierra collected dry branches and broke them into smaller pieces. She then placed the sticks in a fire pi

t and used a match to light the sticks on fire. Next, she propped a metal pan of water over the fire and heated it until the water boiled. Which list correctly describes physical and chemical changes that occurred in this scenario? physical change: breaking sticks chemical changes: lighting the fire, boiling water physical change: boiling water chemical change: breaking sticks, lighting the fire physical changes: lighting the fire, boiling water chemical change: breaking sticks physical changes: breaking sticks, boiling water chemical change: lighting the fire
Physics
1 answer:
tester [92]3 years ago
5 0
The correct option is this: PHYSICAL CHANGES: BREAKING THE STICKS AND BOILING THE WATER. CHEMICAL CHANGE: LIGHTNING THE FIRE.
In chemistry, a physical change refer to a change in which no new material is formed while a chemical change refers to a change in which a new material is formed. Boiling of water does not change water into another thing, even when the water changes to vapor, it will still turn to water again when it cools down, so boiling is a physical change, so also the breaking of sticks. Lightening a fire on the other hand is a chemical change because a new material is formed and the original material can not be recovered. Putting fire on the sticks will turn the sticks to ashes and the sticks can not be recovered from the ashes, so this is a chemical change.
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Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
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Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

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Answer:

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Explanation:

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Using formula of energy

The change in potential energy equal to the change in kinetic energy.

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Put the value into the formula

270\times10^{-3}\times9.8\times50\times10^{-2}=\dfrac{1}{2}\times k\times(24\times10^{-2})^2

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k=45.94\ N/m

Hence, The spring constant is 45.94 N/m.

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