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Dafna11 [192]
3 years ago
12

A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc

h the box vertically into the air. What is the maximum height the box will reach if the spring constant is 300 N/m?
Physics
1 answer:
german3 years ago
6 0

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

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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

5 0
3 years ago
The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands
Ghella [55]

Answer:

Pressure on both feet will be 500N/m^2  

Explanation:

Weight of the person F = 500 N

Area of foot print A=0.5m^2

Area of both the foot A=2\times 0.5=1m^2

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure P=\frac{F}{A}

P=\frac{500}{1}=500N/m^2

So the pressure on both feet will be 500N/m^2 when person stands on both feet.

7 0
3 years ago
Franny drew a diagram to compare images produced by concave and convex lenses.
yanalaym [24]

Answer:

virtual

Explanation:

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Scientists study how the continents move. Why might scientists use a model
tigry1 [53]

Answer:

B. It is too slow to observe directly

Explanation:

They move too slow to be able to observe how they move.


I hope it helps! Have a great day!
bren~

3 0
2 years ago
A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res
VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

KE = PE

\frac{1}{2} mv^2 = mgh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(2(55.8*10^{-2}))}

v = 4.67m/s

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

4 0
3 years ago
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