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2 years ago

11

Calculate the force constant (in N/m) of its plunger's spring if you must compress it 0.160 m to drive the 0.0540 kg plunger to

a top speed of 25.0 m/s.

1 answer:

Leni [432]2 years ago

7 0

Answer:1.318 kN/m

Explanation:

Given

compression $x=0.16 m$

mass of Plunger $m=0.0540 kg$

Top speed $v_{max}=25 m/s$

Conserving Energy

Elastic Potential Energy of spring stored converted in kinetic energy of Plunger

$\frac{1}{2}kx^2=\frac{1}{2}mv^2$

$k\times (0.16)^2=0.054\times 25^2$

$k=1318.35 N/m$

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The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

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3 years ago

Sunshine Corporation is expected to pay a dividend of $1.50 in the upcoming year. Dividends are expected to grow at the rate of

grigory [225]

Answer:

$25

Explanation:

The computation of the intrinsic value of the stock is shown below:

As we know that

Intrinsic value or Price = Dividend ÷ (Cost of equity - growth rate)

where,

Cost of equity or ke is

= Risk free rate of return + beta × (Market rate of return - risk free rate of return)

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So the intrinsic value is

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3 years ago

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Lets do

We know

The rate of change of velocity is acceleration .

$\\ \sf\longmapsto a=\dfrac{dv}{dt}$

$\\ \sf\longmapsto dv=adt$

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$\\ \sf\longmapsto \int dv=a\int dt$

As acceleration is constant .Take it outside of integral .On velocity we can take limit u to v and time from 0 to t

$\\ \sf\longmapsto {\displaystyle{\int}}^v_u dv=a{\displaystyle{\int}}^t_0 dt$

Hence

$\\ \sf\longmapsto v{\huge{|}}^v_u=at$

$\\ \sf\longmapsto v-u=at$

$\\ \sf\longmapsto v=u+at$

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2 years ago

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What does the symbol "E3" represent? ...?

LiRa [457]

The symbol "E3" represents the energy of electrons in the third energy level.

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3 years ago

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A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at

wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

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$t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s$

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

$t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 = \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s$

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

$t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s$

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s - 2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

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