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stellarik [79]
3 years ago
14

An object is attached to a vertical spring and bobs up and down between points A and B. Where is the object located when its kin

etic energy is a minimum?
a. at either A or B
b. midway between A and B
c. one-third of the way between A and B
d. one-fourth of the way between A and B
Physics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

a. at either A or B

Explanation:

Kinetic energy may be defined as the energy of the system or an object which is due to its velocity of the object it possess.

In the context, an object having mass  is attached to spring which is vertical and the object moves up and down due to spring effect between points A and B. Now these points A and B are the extreme points after which the object bounces back.

At point A and B, the velocity of the object becomes zero and hence the kinetic energy of a body varies directly proportional to its velocity.

                  i.e. Kinetic energy $= \frac{1}{2} \text{mass} \times (\text{ velocity})^2$  

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"We the people of the United States, in order to form a more perfect union, establish justice, insure domestic tranquility, prov
sergey [27]

The correct answer is B) Giving Congress the power to create courts

Explanation:

One of the principles the Constitution mentions in the preamble is Justice, this is mentioned in "We the people of the United States, in order to form a more perfect union, establish justice..." According to this, a priority is to make the country have justice, which is possible through courts, laws, among others. In this context, the action that would be allowed is "Giving Congress the power to create courts" because this might ensure the justice principle is applied in the U.S.

4 0
3 years ago
What energy change is associated with the reaction to obtain one mole of H2 from one mole of water vapor? The balanced equation
Yakvenalex [24]

Answer:

ΔH = 249 kJ/mol

Explanation:

The balanced reaction is:

2H₂O(g) → 2H₂(g) + O₂(g)    (1)

To calculate the energy change to obtain one mole of H₂(g) from one mole of H₂O(g), the coefficients of the reaction (1) must be halved:

H₂O(g) → H₂(g) + 1/2O₂(g)    (2)

The enthalpy of the reaction (2) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}  

<em>Where \Delta H_{r}: is the bond enthalpy of reactants and \Delta H_{p}: is the bond enthalpy of products.</em>

<u>For the reactants we have the next bond energies:</u>

2 x (H-O) = 2 x (467)

<u>And the bond energies for the products are:</u>

H-H + (1/2) (O=O) =  436 + (1/2)(498)

So, the enthalpy of the reaction (2) is:

\Delta H = 2 \cdot 467 kJ/mol - 436 kJ/mol - \frac{1}{2} \cdot 498 kJ/mol = 249 kJ/mol  

I hope it helps you!    

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The correct answer Temporary charge
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