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-Dominant- [34]
3 years ago
5

A reaction that releases heat is: endothermic exothermic none of the above

Physics
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:

Exothermic reaction

Explanation:

  • A reaction that releases heat is known as an exothermic reaction.
  • Reactions can be classified as either exothermic or endothermic depending on whether they release or absorb heat from the surrounding.
  • A reaction that absorbs or gains heat from the surrounding is known as endothermic reaction.
  • On the other hand, if a reaction releases heat to the surrounding is known as exothermic reaction.
  • An exothermic reactions results to an increase in temperature of a solution while an endothermic reaction results to a decrease in the temperature of a solution.
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A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
Llana [10]

Answer:

a) C = 40.138\,pF, b) q = 16.056\,nC, c) U = 3.212\,\mu J

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

C = K\cdot \epsilon_{o}\cdot \frac{A}{d}

Where:

K - Dielectric constant.

\epsilon_{o} - Vaccum permitivity.

A - Plate area.

d - Distance between plates.

Hence, the capacitance of the system is:

C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)

C = 4.014\cdot 10^{-11}\,F

C = 40.138\,pF

b) The charge can be found by using the definition of capacitance:

q = C\cdot V_{batt}

q = (4.014\times 10^{-11}\,F)\cdot (400\,V)

q = 1.606\times 10^{-8}\,C

q = 16.056\,nC

c) The energy stored in the charged capacitor is:

U=\frac{1}{2}\cdot Q\cdot V_{batt}

U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)

U = 3.212\times 10^{-6}\,J

U = 3.212\,\mu J

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2 years ago
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