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Alex17521 [72]
3 years ago
12

What type of reaction is being shown in this energy diagram?

Physics
1 answer:
Ratling [72]3 years ago
8 0

Answer:

O exothermic, because energy is released into the surrounding

Explanation:

From the diagram the energy of the reactant is higher than the energy of the product, thereby making it exothermic. If you study diagram well, exothermic reaction means that the reactions releases energy into the surroundings.

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A ramp leading to the entrance of a building is inclied upward at an angle of 7 degree. A suitcase is to be pulled up the ramp b
photoshop1234 [79]

Explanation:

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / F

F = 50 / cos 28°

F = 56.6 lbs

8 0
3 years ago
Suppose the force of gravity on the electron was comparable to the electric force created by the deflection plates. how would th
Alexandra [31]
"F=Vector Sum Of The Two Forces" Is the answer.
3 0
3 years ago
A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu
ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}

Where:

\gamma = \frac{c_{p}}{c_{v}}

\gamma = 1 + \frac{R}{c_{v}}

Final pressure is:

P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}}  \right)^{\gamma}

P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}

P_{3} = \frac{P_{o}}{2^{\gamma}}

8 0
3 years ago
A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?
julia-pushkina [17]
Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

b = mg/v^2 = (0.0023×9.81)/(9.4^2)

b = 0.000255


6 0
2 years ago
In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel
slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

v=60\times \dfrac{1609.34}{3600}=26.822\ m/s

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s

or

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s

The time taken is 2.73414 seconds

The potential energy is given by

U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J

The change in potential energy is 467622.66798 J

5 0
3 years ago
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