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2 years ago

2. A powerful experimental sewing machine is powered by a mass-spring system. This

Physics

1 answer:

Alexus [3.1K]2 years ago

7 0

We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of oscillation motion</h3>

Question Parameters:

This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.

If the <em>sewing </em>machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

Generally the equation for the Time t is mathematically given as

$T=2\pi\sqrt{\frac{m}{k}}$

Therefore

$0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}$

m=2.033e-5kg

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Monochromatic light with wavelength 590 nm passes through a single slit 2. 30 ?m wide and 1. 90 m from a screen. Find the distan

aivan3 [116]

Answer:

The fringes are 4.7*10^-7 m apart, such that they are adjacent.

Explanation:

Using the formula for adjacent fringes given a single slit:

Δ $x=\frac{(Wavelength)(Distance between slit and screen)}{Width}$

Δ $x=\frac{(590/10^{9})(1.90) }{(2.30)}$

Δ $x=0.000000487 m$

Hope this helps!

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2 years ago

Read 2 more answers

What is the change in entropy per mole associated with the melting of gold? the melting point of gold is 1337k and the â†hfus is

riadik2000 [5.3K]

The entropy change<span> of the surroundings is driven by heat flow and the heat flow determines the sign of ΔS</span>surr<span>. It can be calculated by the following expression:

</span>ΔSsurr = -(ΔH) / T

We calculate as follows:

ΔSsurr = -13200 / 1337 = 9.87 J/ K mol

Hope this answers the question. Have a nice day.

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3 years ago

A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?

Advocard [28]

Answer:

$v=697.2km/h$

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

$V=\frac{d}{t}$

The distance is clearly 1743 km and the time is:

$t=2h+30min*\frac{1h}{60min} =2.5h$

Thus, the velocity turns out:

$v=\frac{1743km}{2.5h}\\ \\v=697.2km/h$

Which is a typical velocity for a plane to allow it be stable when flying.

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3 years ago

Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car

lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

$A_a(t) = 5m/s^2$

To get the velocity, we integrate over time:

$V_a(t) = (5m/s^2)*t + V_0$

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

$V_a(t) = (5m/s^2)*t$

To get the position equation we integrate again over time:

$P_a(t) = 0.5*(5m/s^2)*t^2 + P_0$

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

$P_a(t) = 0.5*(5m/s^2)*t^2$

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

$V_b(t) =20m/s$

To get the position equation, we can integrate:

$P_b(t) = (20m/s)*t + P_0$

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

$P_b(t) = (20m/s)*t + 100m$

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

$P_a(t) = P_b(t)$

We can solve this for t.

$0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0$

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

$t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}$

We only care for the positive solution, which is:

$t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s$

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

$P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m$

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4) What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

$V_a(t) = (5m/s^2)*11.48s = 57.4 m/s$

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3 years ago

The initial concentration of acid ha in solution is 0.39 m. if the ph of the solution at equilibrium is 0.76, what is the percen

Alexus [3.1K]

The percent ionization of the acid is 44.56%

<h3>How can we calculate the percent ionization of the acid?</h3>

To calculate the percent ionization of the acid we are using the formula,

The H⁺ ion concentration [H⁺] = C x,

where, we are given,

C= concentration of the acid.

=0.39 M

x= degree of dissociation of the acid.

And one more thing we are given that, the pH of the acid=0.76.

So from the above statement we can say that,

pH = - log [H⁺]

Or,0.76 = -log [H⁺]

Or, log [H⁺] = -0.76

Or, [H⁺] = antilog -0.76

Or,[H⁺]= 10^-0.76

Or,[H⁺]=0.1738.

Now from the above calculation we know, the H⁺ ion concentration= 0.1738 M.

Now we put the known values in the above equation,

[H+]= Cx

Or,0.1739= 0.39 x

Or, x= 0.4459

From the above calculation we can conclude that the percent Ionization of the acid= 0.4459 X 100= 44.59%≈45%

Learn more about Ionization:

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4 0

1 year ago