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Sliva [168]
2 years ago
15

Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work

is needed to stretch the spring from 36 cm to 41 cm
Physics
1 answer:
dlinn [17]2 years ago
4 0

Answer:

0.83 J of work

Explanation:

2 J of work is required to stretch a spring from 34cm to 46cm

So that is 12cm stretched with 2 J of work

We can make that 6cm for 1 J of work

So, we need the find the work for stretching 36cm to 41cm

Which is 5cm

So, What is the work required to stretch 5cm?

1 J of work for 6cm

x work for 5cm

So, by proportion method

1 : 6 :: x : 5

6 * x = 1 * 5

6x = 5

x = 5/6

= 0.83

So to stretch 36cm to 41cm we need 0.83 J of work

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Oh but they are !

Newton's 3rd law of motion says that for every action, the <em><u>re</u></em>action is
equal and opposite.  That's as balanced as you can get.


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3 years ago
Which items do NOT describe possible solutions to air pollution?
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Answer:

2, 6

Explanation:

2 because if you cut down more trees you will have less items to help collect co2

6 because if you have more manufracturing more gasses will be release and moe carbon dioxcide in the air so it will slowly kill the ozone layer.

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Question 4
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6 0
3 years ago
Which of the following statements is true?
fenix001 [56]

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Explanation:

Electromagnetic waves doesnt need to medium for travell

Ex ; Ultra violet rays , X rays

Sound waves needs to medium to travell but electromagnetic waves doesnt

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4 0
3 years ago
A skier moves down a 12° slope at constant speed what can you say about the coefficient of friction assume the speed is low enou
BARSIC [14]

The coefficient of friction is 0.213

Explanation:

We start by writing the equation of the forces along the directions parallel and perpendicular to the incline. We have:

Perpendicular direction:

N-mg cos \theta =0 (1)

where

N is the normal reaction of the plane

mg cos \theta is the component of the weight perpendicular to the plane, with

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=12^{\circ}  is the angle of the incline

Parallel direction:

mg sin \theta - \mu_k N = ma (2)

where :

mg sin \theta is the component of the weight parallel to the slope

\mu_k N is the force of friction, where

\mu_k is the coefficient of friction

N is the normal reaction  

a is the acceleration

In this case, the skier is moving at constant speed, so its acceleration is zero and the equation becomes:

mg sin \theta - \mu_k N =0 (2)

From (1) we find

N=mg cos \theta

And substituting into (2)

mg sin \theta - \mu_k mg cos \theta = 0

Solving for \mu_k,

\mu_k = tan \theta = tan(12)=0.213

Learn more about slope and force of friction here:

brainly.com/question/5884009

#LearnwithBrainly

8 0
3 years ago
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