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makvit
12 days ago
15

Real world trampolines lose energy since they are damped springs with much internal friction. How much energy does the sumo wres

tler lose on each bounce in this situation? b) How can a gymnast keep a constant bounce height in a real world trampoline?
Physics
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In the reaction at Blood Falls, iron and oxygen combine to form iron oxide, which is called rust (water is also present). The re
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A reactant is any substance that takes part in a chemical reaction. Conversely, a product is what emerges from the chemical transformation. A familiar example of a chemical change is rust formation. In this reaction, oxygen and iron, which serve as the reactants, react to produce a substance known as iron oxide, or rust.
3 0
2 months ago
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A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t
Keith_Richards [3271]

Response:

To find power, we must first determine the work done by the force.

1) We will use the following equation to calculate work:

\int\limits {F} \, dx

The force is provided by the problem; our goal is to express 'dx' in terms of 't'

2) It's known that:

\frac{dV}{dt} = a = 2.6

Thus, we have:

v = 2.6t

Then:

\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt

3) Finally, substituting all known values gives us:

\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt

After some calculations, the resulting work is:

161.9638 J.

4) To find power, we will use the following equation:

P = \frac{W}{t}

Thus

P = 161.9638/4.7 = 34.46 W

8 0
2 months ago
A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in
Maru [3345]

Answer:

Explanation:

The transitions occur as follows:

P₁ V₁ changes to 3P₁, V₁ (with constant volume) — first phase.

Subsequently, 3P₁,V₁ transitions to 3P₁, 5V₁ (with constant pressure) — second phase.

During the initial phase, the temperature must be escalated by a factor of 3. Therefore, if the starting temperature is T₁, then the ending temperature will be 3 T₁.

P₁V₁ = n R T₁, where n represents the number of moles of gas.

Thus, nRT₁ = P₁V₁.

The heat added at constant volume is given by n Cv (3T₁ - T₁),

= n x 5/3 R X 2T₁ (noting that for diatomic gas, Cv = 5/3 R).

= 10/3 x nRT₁

= 10/3 x P₁V₁.

In the second phase, the temperature must rise 5 times. Thus, if the initial temperature is 3T₁, then the final temperature will be 15 T₁.

The heat added at constant pressure in this scenario becomes:

= n Cp (15T₁ - 3T₁)

= n x 7/3 R X 12T₁ (for diatomic gases, Cp = 7/3 R).

= 28 x nRT₁

= 28 P₁V₁.

6 0
3 months ago
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