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2 years ago

10

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

, what is the velocity of the sandbag?

1 answer:

marishachu [46]2 years ago

3 0

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

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Write down the principal of lever

pashok25 [27]

Answer:

See the explanation below.

Explanation:

A lever is a simple machine that changes the magnitude and direction of the force applied to move an object. Minimizes the force needed to lift the object.

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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

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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,

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

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

$\Delta H=-310kJ$

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Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

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The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

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$u_2 = -20 m/s$ is the initial velocity of the 2nd puck

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Re-arranging the equation and solving for v, we find:

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