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2 years ago

When a wave strikes and object and bounces off . ( example echo)

Physics

1 answer:

zzz [600]2 years ago

5 0

Answer:

C. reflection

Explanation:

reflection occurs when a wave strikes an object and bounces off of it

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A train moves at constant velocity of 90km/h. How far will it go in 0.25h.

enot [183]

In a quarter of an hour it will travel 1/4 x 90 = 22.5Km

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2 years ago

An electric field can be modeled as lines of force radiating out into space.

Hatshy [7]

Yes. Electric field lines are the imaginary lines that represents strength & direction of electric field due to a charge. It can be modeled as lines.

In short, Your Answer would be "True"

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2 years ago

Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in

Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

$4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $X_2+3Y_2\rightarrow 2XY_3$ $\Delta H_1=-370kJ$

(2) $X_2+2Z_2\rightarrow 2XZ_2$ $\Delta H_2=-120kJ$

(3) $2Y_2+Z_2\rightarrow 2Y_2Z$ $\Delta H_3=-270kJ$

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) $4XY_3\rightarrow 2X_2+6Y_2$ $\Delta H_1=2\times (+370kJ)=740kJ$

(2) $2X_2+4Z_2\rightarrow 4XZ_2$ $\Delta H_2=2\times (-120kJ)=-240kJ$

(3) $6Y_2+3Z_2\rightarrow 6Y_2Z$ $\Delta H_3=3\times (-270kJ)=-810kJ$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+740kJ)+(-240kJ)+(-810kJ)$

$\Delta H=-310kJ$

Therefore, the change in enthalpy of the reaction is, -310 kJ

5 0

3 years ago

The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th

worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

$\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg$

where $m_a$ is the mass of the arrow, $m_b$ is the mass of the block, $\Delta H$ of the change in height of the block after the collision, and $v$ is the velocity of the arrow before it hit the block.