Physics, 06.10.2019 11:00 mercedes73

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

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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

Answer from: tigistamare03

We have to heat a brass rod at 982.4 $\rm ^\circ C$ for it to be 1.8 % longer than it is at 30.

Given :

Brass rod is 1.8 % longer than it is at 30 $\rm ^\circ C$ .

Solution :

We know that,

$\rm \dfrac{\Delta L}{L_0}=\alpha \Delta T$

$\rm \dfrac{\Delta L}{L_0}=\alpha ( T_2-T_1)$ ---- (1)

Where, $\rm \Delta L$ is the elongation,

L_0 is the original length,

$\alpha$ is the coefficient of linear expansion. For brass,

$\rm \alpha = 18.9\times 10^-^6/^\circ C$ ,

$\rm \Delta T$ is the change in temperature.

1.8 % length expansion means:

$\rm \dfrac{\Delta L}{L_0} = 0.018$

Now put the values of

$\rm \alpha ,\;\dfrac{\Delta L}{L_0},\;and\;T_1$ in equation (1) we get:

$\rm 0.018 = 18.9\times 10^-^6 \times(T_2 - 30)$

$\rm T_2 = 982.4\; ^\circ C$

We have to heat a brass rod at 982.4 $\rm ^\circ C$ for it to be 1.8 % longer than it is at 30.

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Answer from: Quest

True.it is better to buy a washer with a higher efficiency rating

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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