Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.

Answers

Answer 1

Answer:

Answer:

av=0.333m/s, U=3.3466J

$v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_(B2)-v_(A2)$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

$2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s$

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Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer:

$\dot W_(in) = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_(u)\cdot T$

$P\cdot V = (m)/(M)\cdot R_(u)\cdot T$

$\rho = (P\cdot M)/(R_(u)\cdot T)$

$\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))$

$\rho = 1.2\,(kg)/(m^(3))$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )$

$\dot m = 0.18\,(kg)/(s)$

The work input is:

$\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)$

$\dot W_(in) = 49.386\,kW$

If the rise and fall of your lungs is considered to be simple harmonic motion, how would you relate the period of the motion to your breathing rate (breaths per minute)? Breaths per minute is an angular frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is its reciprocal. Breaths per minute is an angular frequency. The period is its reciprocal.

Answers

Answer:

Breaths per minute is a frequency. The period is its reciprocal.

Explanation:

In simple harmonic motion, a period (T) is the time taken for one point to start in a position and reach that position again, in other words to complete a cycle or lapse. In this case, a period is the time one takes from starting to inspire the air to releasing all of it from the lungs.

In simple harmonic motion, the frequency (f) is how many times a point completes a cycle or lapse in one unity of time (could be one second, one minute, one hour, etc). In this case, the frequency is how many times one breathes in one minute. This is the breathing rate, since it is breathings per minute. Breaths per minute is a frequency.

Period (T) and frequency (f) relate to each other in the following formulae: $T=(1)/(f)$ or $f=(1)/(T)$ .

Therefore, breaths per minute is a frequency, and since it is related to the period, we say the period is reciprocal to it.

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?

Answers

Answer:

$(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = (d)/(2)$

$r = (0.03)/(2)$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^(-4 ) \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o( d \phi )/(dt )$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

$\mu_o I + \epsilon_o \mu _o( d \phi )/(dt ) = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^(-12 ) \ m^(-3) \cdot kg^(-1)\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

$\mu_o I + \epsilon_o \mu _o ( d [EA] )/(dt ) = 0$

=> $(dE)/(dt) =- (I)/(\epsilon_o * A )$

=> $(dE)/(dt) =- (17)/(8.85*10^(-12) * 7.07*10^(-4) )$

=> $(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

What kind of energy is produce when sun reaches solar panel?

Answers

Answer:

Radient to ElEcTrIcAAl

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

The energy produced when the sun reaches solar panel is nuclear fusion

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.

Answers

Answer:

The answer is 9 m.

Explanation:

Using the kinematic equation for an object in free fall:

$y = y_o - v_o-(1)/(2)gt^(2)$

In this case:

$v_o = \textrm{Initial velocity} = 1.5[m/s]\nt = \textrm{air time} = 1.2 [s]$ $y_o = 0$

$g = \textrm{gravity} = 9.8 [m/s^(2) ]$

Plugging those values into the previous equation:

$y = 0 - 1.5*1.2-(1)/(2)*9.8*1.2^(2) \ny = -8.85 [m] \approx -9 [m]$

The negative sign is because the reference taken. If I see everything from the rescuer point of view.

A solid cylinder of cortical bone has a length of 500mm, diameter of 2cm and a Young’s Modulus of 17.4GPa. Determine the spring constant ‘k’. Please explain.