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goblinko [34]
2 years ago
6

A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the

n changes to k2 = 250 N/m as the spring is stretched to x2 = 65 cm. From x2 = 65 cm to x3 = 89 cm the spring force is constant at F3 = 105 N.
Write an equation for the work done in stretching the spring from x1 to x2.
Calculate the work done, in joules, in stretching the spring from x1 to x2.
Calculate the work, in joules, necessary to stretch the spring from x = 0 to x3.
Physics
1 answer:
pantera1 [17]2 years ago
6 0

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

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soldi70 [24.7K]

Answer:

The concentration downstream reduces to 0.03463mg/L

Explanation:

Initially let us calculate the time it is required to fill the lake, since for that period of time the pollutant shall remain in the lake before being flushed out.

Thus the detention period is calculated as

t=\frac{V}{Q}\\\\t=\frac{9000000}{7}=1.285\times 10^{6}seconds\\\\\therefore t = 14.872days

Now the concentration of the pollutant after 14.872 days is calculated as

N_{t}=N_{0}e^{-kt}

where

N_{o} is the initial concentration

't' is the time elapsed after which the remaining concentration is calculated

k is the dissociation constant.

Applying values we get

N_{t}=3\times e^{-0.3\times 14.872}

N_{t}=0.03463mg/L

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3 years ago
A ball with a mass of 0.5 kg is attached to one end of a light rod that is 0.5 m long. The other end of the rod is loosely pinne
prohojiy [21]

Answer:

The tension in the rod as the ball moves through the bottom circle is 9.8 N

Explanation:

When the ball is released from rest, the centripetal force equals the weight of the ball. So mv²/r = mg where m = mass of ball = 0.5 kg, v = speed of ball, r = radius of vertical circle = length of rod = 0.5 m and g = acceleration due to gravity = 9.8 m/s²

v = √gr = √9.8 m/s² × 0.5 m = √4.9 = 2.21 m/s

Now at the bottom of the circle T - mg = mv²/r where T = tension in the rod

T = m(g + v²/r)

= m(g + (√gr)²/r)

= m(g+ gr/r)

= m(g + g)

= 2mg

= 2 × 0.5 kg × 9.8 m/s²

= 9.8 N

So, the tension in the rod as the ball moves through the bottom circle is 9.8 N

8 0
4 years ago
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8 0
3 years ago
I REALLY REALLY REALLY NEED HELP WITH THIS PLSSSSSS ASAP FIRST ANSWER GETS BRAINLIST
aev [14]

Answer:

I think 80J

Explanation:

50 percent efficient means 50% output

3 0
2 years ago
Read 2 more answers
(pleases show work)
Feliz [49]

a. 46 m/s east

The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:

v=u+at

where

v is the velocity calculated at time t

u is the initial velocity

a is the acceleration

The jet in the problem has, taking east as positive direction:

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

Substituting t = 10 s, we find the final velocity of the jet:

v=16 + (3)(10)=46 m/s

And since the result is positive, the direction is east.

b. 310 m

The displacement of the jet can be found using another suvat equation

s=ut+\frac{1}{2}at^2

where

s is the displacement

u is the initial velocity

a is the acceleration

t is the time

For the jet in this problem,

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

t = 10 s is the time

Substituting into the equation,

s=(16)(10)+\frac{1}{2}(3)(10)^2=310 m

4 0
3 years ago
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