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Sholpan [36]
3 years ago
5

What is efficiency of a machine?

Physics
2 answers:
Nataly_w [17]3 years ago
7 0

Answer:

<em>Efficiency</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>machine </em><em>is</em><em> </em><em>defined </em><em>as</em><em> </em><em>the</em><em> </em><em>ratio</em><em> </em><em>of</em><em> </em><em>output </em><em>work</em><em> </em><em>to</em><em> </em><em>input </em><em>work </em><em>in</em><em> </em><em>a</em><em> </em><em>machine </em><em>.</em><em> </em><em>It</em><em> </em><em>is</em><em> </em><em>expressed</em><em> </em><em>in</em><em> </em><em>percentage </em><em>and</em><em> </em><em>denote</em><em>d</em><em> </em><em>by</em><em> </em>

<em>η</em><em> </em><em>(</em><em> </em><em>eta</em><em>)</em><em>.</em>

Flura [38]3 years ago
6 0
Mechanical efficiency is a measure of how well the machine converts the input work or energy into some useful output. It is calculated by dividing the output work by the input work. The ideal machine has mechanical efficiency equal to unity, while the real machine has mechanical efficiency less than unity
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The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0
3 years ago
Define average velocity.Immersive Reader
vagabundo [1.1K]

Answer:

The slope of a graph of position vs time

4 0
3 years ago
19. State any 3 applications of capillary action (3mk)​
zlopas [31]

Explanation:

1. Movement of water, food and mineral salts in plants

2. Absorption of water by towels when wiping our bodies

3. It is used to absorb ink using a blotting paper or tissue

7 0
3 years ago
Which, if any, of the following statements about electric field lines is/are true? A.The electric field is always perpendicular
jonny [76]
The electric field is always perpendicular to the surface outside of a conductor. TRUE

<span> If an electron were placed on an electric field line, it would move in a direction perpendicular to the field. FALSE, it would move in an anti-parallel direction because its charge is negative </span>
 
<span>Electric field lines originate on positive charge and terminate on negative charge. TRUE ; but they can also go to infinity </span>
 
It is possible for two electric field lines to cross each other.
<span> Usually FALSE; though technically possible at special points where field is zero. </span>
 
If an electron and a positron were in the presence of a very strong electric field, they would move away from each other.
<span> TRUE; one is positive, and one is negative. If the field is strong enough, the action of the field will overcome the mutual attraction between them </span>
 
It is not possible for the electric field to ever be zero. FALSE: it IS possible, inside a conductor for instance
   
If a proton were placed on an electric field line, it would move in a direction anti-parallel to the field.
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8 0
3 years ago
A bicycle racer inflates their tires to 7.1 atm on a warm autumn afternoon when temperatures reached 27 °C. By morning the tempe
natulia [17]

Answer:

The required pressure is 6.4866 atm.

Explanation:

The given data : -

In the afternoon.

Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa =  709.275 Kpa

Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K

In the morning .

Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K

Given that volume remains constant.

To find final pressure ( p₂ ).

Applying the ideal gas equation.

p * v = m * R * T

\frac{p}{T}  = constant

\frac{p_{1} }{T_{1} }  = \frac{p_{2} }{T_{2} }

p_{2}  = \frac{T_{2} }{T_{1} } *p_{1}  

p_{2}  = \frac{278}{300}  * 709.275  = 657.2615 Kpa = 6.486 atm

8 0
2 years ago
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