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anyanavicka [17]
3 years ago
11

Which is not a way to accelerate an object?

Physics
2 answers:
Fittoniya [83]3 years ago
4 0
Slowing it down <span>is not a way to accelerate an object. The answer is letter B. Speeding it up, changing its direction and applying balanced forces are ways to accelerate the object</span>
densk [106]3 years ago
3 0
I would definitely think its B....
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It’s E we just had a test in this and I got it right
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what is the force that opposes the movement of two surfaces that are in contact and are sliding over each other? and it is not f
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Answer:

Friction

Explanation: is the force that opposes motion between any surfaces that are in contact. There are four types of friction: static, sliding, rolling, and fluid friction. Static friction acts on objects when they are resting on a surface.Nov 3, 2021

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2 years ago
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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

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3 years ago
Explain how airplane wings and the wings of birds affect the air flowing over them to produce lift
d1i1m1o1n [39]
Because the air in their wings helps them fly
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