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4 years ago

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Physics

1 answer:

Agata [3.3K]4 years ago

7 0

Answer:

The Answer is A) Metal X aluminium; Metal Y platinum

Explanation:

The behavior of the bimetallic strip depends on the coefficient of thermal expansion. Where the coefficient of the metal X should be greater than the coefficient of thermal expansion for metal Y. In the attached image we can see the effect of this particular behavior.

In the only case where the coefficient of thermal expansion of the metal X is greater than the used in the metal Y is in case A.

Here we have:

$\alpha _{alum}=25*10^{-6} [1/C]\\ \alpha _{plat}=9*10^{-6} [1/C]\\\\alpha _{copp}=17*10^{-6} [1/C]\\\\alpha _{steel}=1*10^{-6} [1/C]\\$

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Define speed.

Vinvika [58]

The answer to the question is choice 2

7 0

3 years ago

At an accident scene on a level road, investigators measure a car's skid mark to be 88 m long. It was a rainy day and the coeffi

german

Answer:

The speed was 26.91m/s (96.9 km/h)

Explanation:

Here you have to consider that at the beginning you have an amount of kinetic energy (K) that is dissipated because of the work done by friction forces (T). Since the car is stopped after the accident, all the energy has been dissipated. Thus, $K=T$ .

The definition of the kinetic energy is $K=1/2*m*v^2$ .

The work done by the friction forces is: $T=f*(m*g)*d$ . Where f is the friction coefficient, g is the gravity acceleration, m is the mass of the car and d is the skid marks longitude. Therefore,

$1.2*m*v^2=f*(m*g)*d$

Since m is in both sides it can be cancelled so it is not necessary to considered.

Then, the speed is determined by the following equation:

$v=\sqrt{2*f*g*d} =\sqrt{2*0.42*9.8m/s^2*88 m}=26.91 m/s$

4 0

3 years ago

Derive an expression for the acceleration of the car. Express your answer in terms of D and vt Determine the time at which the s

Bezzdna [24]

Answer:

V(car) = V(truck) at t = Dt/2

acceleration = v(car) = D/t^2

Explanation:

acceleration = v(car) = D/t^2

Since the average velocities must be the same, the car's final velocity must be twice the trunk velocity assuming the car start with zero velocity, since acceleration remain the same throughout the journey velocities at half-time point must be equal.

3 0

4 years ago

1) A particle move along the x axis, as x=tsquare-2t+1(SI unit). (a) The functions of its velocity and accelerator vector agains

Bond [772]

Answer:

$\text{I believe this is calculus, no? But I believe that there should be a y-component of position}\\\text{But based off of the information provided, here goes:}\\\text{(a)}\\x'(t)=v(t)=\frac{d}{dt}t^2-2t+1\\v(t)=2t-2\text{ is the velocity vector}\\v'(t)=a(t)=\frac{d}{dt}2t-2\\a(t)=2 \text{ is the acceleration vector}\\\text{(b)}\\\text{Average Velocity }=\frac{x_{f}-x{i}}{t_{f}-t_{i}}\\=\frac{1-1}{2}=0, \text{based off of the information provided, the average velocity in the first 2 seconds}: 0m/s$

8 0

3 years ago

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

$P=\frac{W}{t}$

The work is the force the is applicated in a distance so

W=F*d

replacing:

$P=F*\frac{d}{t}$ and $\frac{d}{t}$ determinate distance in time is velocity so

a).

$P=F*v$

$F=m*a\\F=m*g*sin(33.5)$

$P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW$

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

$vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}$

The maximum force is when the car is accelerating so

$Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N$

so the maximum force is the maximum force by the maximum speed

$Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW$

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ

4 0

4 years ago

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