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3 years ago

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A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and

spring B has a spring constant of 5 N/m.
How do the lengths of the springs compare?

A:Spring B is 1 m longer than spring A because 5 – 4 = 1.
B:Spring A is the same length as spring B because 60 – 60 = 0.
C:Spring B is 60 m longer than spring A because 300 – 240 = 60.
D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

1 answer:

goblinko [34]3 years ago

5 0

Answer:

D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

Step-by-step explanation:

In this question, you should remember the Hooke's Law in physics.

The Hooke's Law simply explains that the extension that occurs on a spring is directly proportional to the load applied on it.

The mathematical expression for this law is

$F=-kx$

where;

F= force applied on the spring

x = the extension on the spring

k= the spring constant which varies in spring.

The question will need you to calculate the extension on the springs A and B then compare the values obtained.

<u>In spring A</u>

Force, F=60N and spring constant ,k=4 N/m

To find the extension x apply the expression;

$F=-kx\\\\60=-4*x\\\\60=-4x\\\\\frac{60}{-4} =\frac{-4x}{-4} \\\\\\-15=x$

Here the spring extension is 15 m

<u>In spring B</u>

Force, F=60N and spring constant , k=5N/m

To find the extension x apply the same expression

$F=-kx\\\\60=-5*x\\\\60=-5x\\\\\\\frac{60}{-5} =\frac{-5x}{-5} \\\\\\-12=x$

Here the extension on the spring is 12 m

<u>Compare</u>

The extension on spring A is 3 m longer than that in spring B because when you subtract the value of spring B from that in spring A you get 3m

$=15m-12m=3m$

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salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

$f(x)=\frac{e^{\frac{-x}{1000} }}{1000}$

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

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Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

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P(a<x<b) = $\frac{1000}{1000}.e^{\frac{-x}{1000} }$

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a. For a component to last more than 3000 hours:

P(3000<x<∞) = $e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}$

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = $e^{-3}$

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = $e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}$

P(1000<x<2000) = $e^{-2}-e^{-1}$

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c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = $e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}$

P(0<x<1000) = $e^{-1}-1$

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${\frac{-x}{1000} }=ln0.9$

-x = -1000.ln(0.9)

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10% of the components will have failed at 105.4 hours.

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