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IgorLugansk [536]
3 years ago
12

What is the efficiency of a machine that uses 102kj of enegery to do 98 kJ of work?

Physics
1 answer:
WINSTONCH [101]3 years ago
7 0

Answer: I got an increase of 306 kJ in internal energy.

Explanation: I used the 1st Law and the sign convention to get the answer in the screenshot.

(sorry if i'm wrong)

:(

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The resting heart rate will increase quickly in a person who has a high level of cardiovascular fitness. True or false
mestny [16]
False . It won't increase, take an athlete for example . They don't get tired as easily because their cardiovascular fitness is strong .
6 0
3 years ago
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Which process is represented by the PV diagram shown below?
MrMuchimi

Answer: B. the isovolumetric process

Explanation:

In the graph given, the volume is constant throughout. It represents a constant volume process. Such processes are called the isovolumetric process or isochoric process.

<em>Hence, option B is the correct answer.</em>

Option A is incorrect because in an isobaric process, the pressure is constant.

Option C is incorrect because in an isothermal process, the temperature is constant.

Option D is incorrect because in an adiabatic process there is no heat transfer.

8 0
3 years ago
Juan is making ice tea. When he adds ice to the tea, why does the tea cool down?​
Tresset [83]

Answer: because the ice was cold, and it melted in the tea. So the tea is now cooler.

Explanation:

4 0
3 years ago
When the body requires an increased blood flow rate in a particular organ or muscle, it can accomplish this by increasing the di
horsena [70]

Answer: The percentage increase in diameter is 19%

Explanation:

<em>Using Poiseuille's law which states that Flow rate , Q = ΔPπr⁴/8ηl</em>

<em>where ΔP is pressure difference, π is a constant, r is the radius which is half of the diameter, η is viscosity of blood, l is length of blood vessel</em>

Let Q₁ be the intitial flow rate, Q₂ the final flow rate,r₁ and r₂ the initial and final radius.

Note: r = d/2, therefore, r₁⁴ = d₁⁴/16 r₂⁴ = d₂⁴/16

Also, Q₂ = 2Q₁, since the flow rate is doubled

Since all factors except diameter is constant, d₂⁴/16 = 2d₁⁴/16

d₂⁴/16 =  d₁⁴/8

multiply both sides of the equation by 16

d₂⁴ = 2d₁⁴

taking the fourth root of both sides

d₂ = 1.19*d₁

d₂ - d₁ = 1.19d₁ - d₁

d₂ - d₁ = d₁(1.19 - 1)

d₂ - d₁ = 0.19d₁

d₂ - d₁/d₁ = (0.19d₁/d₁)*100%

(d₂ - d₁/d₁)*100% = 19%

Therefore, the percentage increase in diameter is 19%

3 0
4 years ago
Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble
Llana [10]

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v²  = μRg

= .683  x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

3 0
3 years ago
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