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A sheet of aluminum alloy is cold-rolled 33 percent to a thickness of 0.096 in. If the sheet is then cold-rolled to a final thickness of 0.061 in., what is the total percent cold work done

Answers

Answer 1

Answer:

The total percent cold work done is 36.46%

Explanation:

Let initial metal thickness = T

Final metal thickness = t

The percent cold work done = WC

Then

%Wc = (T - t)/T × 100

% Wc = ( 0.096 - 0.061 )/0.096 ×100

Total %WC = 36.46%

Answer 2

Answer:

Answer:

The total percent of cold work is 57.34%

Explanation:

Let x the initial thickness of the sheet. After 33% of cold working, the thickness is 0.096 in. Then:

x - 0.33x = 0.096

x = 0.143 in

the final thickness is equal to 0.061 in. The percent of cold work done is:

$percent-of-cold-work=((initial-thickness)-(final-thickness))/(initial-thickness)*100$

$percent-of-cold-work=(0.143-0.061)/(0.143) *100=57.34$ %

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As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?

Answers

Answer:

The value of g is $g =76.2 m/s^2$

Explanation:

From the question we are told that

The mass of the weight is $m = 1.30 kg$

The spring constant $k = 1.73 g/m = 1.73 *10^(-3) \ kg/m$

The second harmonic frequency is $f = 100 \ Hz$

The number of oscillation is $N = 200$

The time taken is $t = 315 \ s$

Generally the frequency is mathematically represented as

$f = (v)/(\lambda)$

At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated

$l = \lambda$

Noe from the question the vibrating string is just half of the length of the main string so

Let assume the length of the main string is $L$

So $l = (L)/(2)$

The velocity of the vibrating string is mathematically represented as

$v = \sqrt{(T)/(\mu) }$

Where T is the tension on the string which can be mathematically represented as

$T = mg$

$v = \sqrt{(mg)/(k) }$

Then

$f = (v)/((L)/(2) )$

=> $v = (fL )/(2)$

=> $\sqrt{(mg)/(k) } = (fL)/(2)$

=> $g = (f^2 L^2 \mu)/(4m)$

substituting values

$g = ((100) * (1.73 *10^(-3) ))/((4 * 1.30)) L^2$

$g = 3.326 m^(-1) s^(-2) L^2$

Generally the period of oscillation is mathematically represented as

$T_p = 2 \pi \sqrt{(L)/(g) }$

=> $L = (T^2 g)/(4 \pi ^2)$

The period can be mathematically evaluated as

$T_p = (t)/(N)$

substituting values

$T_p = (315)/(200)$

$T_p = 1.575 \ s$

Therefore

$L = (1.575^2 * g )/(4 \pi ^2)$

$L = 0.0628 ^2 g$

$g = 3.326 m^(-1) s^(-2) L^2$

substituting for L

$g = 3.326 ((0.0628) g)^2$

=> $g = (1)/((3.326)* (0.0628)^2)$

$g =76.2 m/s^2$

A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________

Answers

The speed when all the fuel has been exhausted is 2415m/s

According to this question, the following information was given:

Exhaust velocity of fuel, V(e)= 1500 m/s
Initial speed of rocket, V₁ = 0 m/s
Final speed of rocket, V₂ = ?
Fuel weight = 80% of total weight

Using Tsiolkovsky rocket equation as follows:

∆V = V(e) ln(m1/m2)

m1 = initial mass
m2 = final mass without repellant

m2 = m1 - 80%m1

m2 = m1 - 0.8m1

m2 = 0.2m1

∆V = V2 - V1

Hence;

V2 - 0 = 1500 × ln (m1/0.2m1)

V2 = 1500 ln(1/0.2)

V2 = 1500 × 1.609

V2 = 2415m/s.

Therefore, the speed when all the fuel has been exhausted is 2415m/s.

Learn more at: brainly.com/question/19531823?referrer=searchResults

Answer:

$v_2 =2414\ m/s$

Explanation:

given,

exhaust velocity of fuel(v_e) = 1500 m/s

initial speed of rocket,v₁ = 0 m/s

final speed of rocket, v₂ = ?

fuel weigh = 80 % of total weight

using Tsiolkovsky rocket equation

$\Delta v = v_e ln((m_1)/(m_2))$

Δ v = v₂ - v₁

v_e is the exhaust speed

m₁ is the initial total mass.

m₂ is the is the final total mass without propellant.

m₂ = m₁ - 0.8 m₁

m₂ = 0.2 m₁

$v_2-v_1 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 =2414\ m/s$

When all the fuel is exhausted speed of the fuel is equal to $v_2 =2414\ m/s$

A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?

Answers

Answer:

$R\approx12.43 \,\, \Omega$

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

$V=I\,\,R\nR = (V)/(I) \nR=(87)/(7) \nR\approx12.43 \,\, \Omega$

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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Define the term energy density of a body under strain

Answers

Answer:

Please mark as Brainliest!!

Explanation:

Strain energy is defined as the energy stored in a body due to deformation. The strain energy per unit volume is known as strain energy density and the area under the stress-strain curve towards the point of deformation. When the applied force is released, the whole system returns to its original shape.

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answers

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=(GMm)/(r^2)$

Where $G=6.67*10^(-11)Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=(GMm)/(r^2)$

Which means:

$a=(GM)/(r^2)$

The object departs from rest ( $v_0=0m/s$ ) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$ , which for our case will be:

$v^2=2ad=(2GMd)/(r^2)$

$v=\sqrt{(2GMd)/(r^2)}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5*10^3m$ . With our values then we have:

$v=\sqrt{(2GMd)/(r^2)}=\sqrt{(2(6.67*10^(-11)Nm^2/Kg^2)(2*10^(30)Kg)(0.025m))/((5*10^3m)^2)}=516526.9m/s$

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.91 m/s (96.00 mph) . Assuming a pitched ball has a mass of 0.1434 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?

Answers

Answer: 132.02 J

Explanation:

By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = 132.02 J

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