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2 years ago

What is the least number of forces required to stretch a spring?

Physics

1 answer:

Solnce55 [7]2 years ago

7 0

Answer:

The least number of forces required to stretch a spring is one.

Explanation:

Let suppose that spring is ideal, that is, that effects from its mass can be neglected since it is insignificant in comparison with external forces. In addition, let the spring have a linear behavior, meaning that net external longitudinal force exerted on spring is directly proportional to defomation. (Hooke's Law) That is:

$F \propto x$ (1)

$F = k\cdot x$ (2)

Where:

$F$ - Net external force, measured in newtons.

$k$ - Spring constant, measured in newtons per meter.

$x$ - Deformation of spring, measured in meters.

Hence, the least number of forces required to stretch a spring is one.

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Un libro del peso di 12 N è in equilibrio su un tavolo. Sapendo che il coefficiente di attrito statico vale 0,5, la forza di att

Tanzania [10]

Answer:

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

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3 years ago

A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A

Rashid [163]

A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

$IMA = \frac{d_i}{d_o}$

where

$d_i$ is the input distance

$d_o$ is the output distance

For the pulley system in this problem, $d_i = 3.90 m$ and $d_o = 0.975 m$ , so the IMA is

$IMA=\frac{3.90 m}{0.975 m}=4$

B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

$MA=\frac{F_o}{F_i}$

where $F_o$ is the output force and $F_i$ is the input force. For the pulley system in this problem, $F_i = 375 N$ and $F_o = 1345 N$ , so the MA is

$MA=\frac{1345 N}{375 N}=3.59$

C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

$\eta = \frac{MA}{AMA} \cdot 100$

Therefore, in this case,

$\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%$

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3 years ago

What is the displacement from a starting position of (14.0, 3.0) m to a final position of (−3.0, −4.0) m?

dem82 [27]

Answer:

change in y = -7

change in x = -17

magnitude of displacement = sqrt(7^2+17^2)

tan of angle below -x axis = 7/17

because in third quadrant where x and y are negative

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3 years ago

A bullet travels at 850 m/s. how long will it take a bullet to go 1 km?

Dmitrij [34]

The speed of bullet =
850 m/s

Distance given = 1 km = 1000m

S = D/t
t • S = D/t • t
St = D
St/S = D/S
t = D/S
t = 1000m/850m/s
t = 1.176 s

It will take the bullet 1.176 or about 1.18 seconds to go 1 km.

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3 years ago

When force is applied to a breaker bar the torque can be calculated by multiplying the length of the lever by the?

Nimfa-mama [501]

When a force applied to a breaker bar the torque can be calculated by multiplying the<u> length of the lever</u> by the tangential component of force on the lever.

<h3>What is torque?</h3>

Torque is the <u>rotating equivalent</u> of force in physics and mechanics. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.

Torque is given by the formula :

α = r x F ( bold letters represent vector quantities)

The S.I. unit for torque is : N - m ( Newton - meter)

<h3>How do we define 1 N-m of torque?</h3>

The newton-metre is a torque unit (also known as a moment) in the SI system. The torque produced by a one newton force applied <u>perpendicularly to the end of a one metre long</u> moment arm is known as a newton-metre.

To learn more about torque:

brainly.com/question/14970645

#SPJ4

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2 years ago