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3 years ago

To what temperature would you have to heat a brass rod for it to be 1.8 % longer than it is at 30 ∘c?

1 answer:

3 0

In physics, certain metals elongate when it is heated. This is a consequence of the expansion of the molecules present in a metal tube, for example. This elongation is described by the equation:

ΔL = L0*α*ΔT, where

ΔL is the elongation. In other words, this is the difference between the original length and the elongated length.
L0 is the original leng
α is the coefficient of linear expansion. This is an empirical data for specific kind of materials. For brass, α = 18.9 x 10^6/°C
ΔT is the change in temperature

Rearranging the equation,

ΔL/L0 = α*ΔT, where ΔL/L0 is the percentage of length expansion which is equal to 0.018 (1.8^%)

0.018 = (18.9 x 10^-6)(T-30)
T = 982.4°C

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3 years ago

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A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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3 years ago

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5

Leto [7]

Answer:

The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.

Explanation:

ω1= 1.72x10^4 rad/sec

ω2= 5.42x10^4 rad/sec

ωmax= 8.42x10^4 rad/sec

θ= 1.72x10^4 rad

$\alpha = \frac{w2^{2}-w1^{2} }{2*(\theta2 - \theta1)}$

α=7.67 x10^4 rad/sec²

t= ωmax / α

t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²

t=1.097 sec

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3 years ago

Question 3 of 10

Mazyrski [523]

Answer:

B . energy cannot be created or destroyed

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3 years ago

A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire), Fig. 5–39.

Trava [24]

Answer:

1742.24106 revolutions per day

Explanation:

v = Velocity

d = Diameter = 1.1 km

r = Radius = $\dfrac{d}{2}=\dfrac{1.1}{2}=0.55\ km$

g = Acceleration due to gravity = 9.81 m/s²

g = 0.9 g

The centrifugal force will balance the gravitational force

$F_c=mg\\\Rightarrow \dfrac{mv^2}{r}=m0.9g\\\Rightarrow v=\sqrt{\dfrac{0.9gmr}{m}}\\\Rightarrow v=\sqrt{0.9gr}\\\Rightarrow v=\sqrt{0.9\times 9.81\times 0.55\times 10^3}\\\Rightarrow v=69.68464\ m/s$

$\dfrac{1}{T}=\dfrac{v}{2\pi r}\\\Rightarrow \dfrac{1}{T}=\dfrac{69.68464}{2\pi 0.55\times 10^3}\times 24\times 60\times 60\\\Rightarrow \dfrac{1}{T}=1742.24106\ rev/day$

The rotation speed is 1742.24106 revolutions per day

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3 years ago

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