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alex41 [277]
2 years ago
5

for a certain chemical reaction, the reactants contain 52 kj of potential energy and the products contain 32 kj how much energy

is absorbed or released by the reaction. A.84 kj is released B.20 kj is released C. 20 kj is absorbed D.84 kj is absorbed
Physics
2 answers:
Luden [163]2 years ago
7 0
B, is the answer. (20 kj is released)
mojhsa [17]2 years ago
4 0

Answer:

B. 20 kj is released

Explanation:

A.P.E.X. CST

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A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the
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As per the question, the mass of meteorite [ m]= 50 kg

                       The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

                                       Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2

                                                             =\frac{1}{2}50kg*[1000\ m/s]^2

                                                               =\frac{1}{2}50* 10^{6}\ J

                                                               =25*10^6\ J    

Here, J stands for Joule which is the S.I unit of energy.

Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J

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In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

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The object is at rest is the answer.
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