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dangina [55]
3 years ago
12

An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is corr

ect at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What wull be the length of rod at 0 degree celcius?​
Physics
2 answers:
natulia [17]3 years ago
5 0
Given:
temperature at which the rod is measured= 25°
length of aluminium rod measured at 25°C= 1 meter
temperature at which the scale is correct= 0°C
Linear expansivity of steel= 12* 10^-6
linear expansivity of aluminum= 26* 10^-6
To find:
True length of rod at 25°C
Length of rod at 0°C
Solution:
Let the length of the aluminum and steel rod at 0°C be L0A and L0S respectively.
Length of the steel scale at 25°C , thermal expansion is:
LA25 = LA0(1+ (αS - αA)25)
LA0 = LA25/ (1+ (αS-αA)t)
LA0 = 100/ (1+ (12*10^-6 - 26*10^-6)25)
LA0 ≈ 100.05 cm at 0°C
2. Let the length at 25°C of the aluminium trod be LA25
The actual measure of aluminum rod at 25°C, will be
LA25 = 100 * (1+ 26* 10^-6*25)
LA25 ≈ 100.06 cm
marin [14]3 years ago
3 0

Answer:

An almunium rod when measured with a steel scale,both being at 25°C appears to be 1meter long. If the scale is correct at 0°C

Explanation:

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A 5.0-kg bar of lead is placed inside a 12-L chamber filled with helium gas. The temperature of the lead and helium is the same.
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As the temperature of the lead and helium is the same. Thus the average kinetic energy is also the same for lead and helium.

Reason:

It is given that a 5.0-kg bar of lead is placed inside a 12-L chamber filled with helium gas. The temperature of the lead and helium is the same. It is required to compare the average kinetic energy of the lead atoms and helium atoms.

The average kinetic energy is calculated as, K=\frac{3}{2} \frac{R}{N} T.

Here K is the average kinetic energy,  R is the gas constant, N is the Avogadro's number, and T is the temperature.

As the temperature is the same for both lead and helium. As a result, the average kinetic energy is also the same for lead and helium.

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2 years ago
Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo
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Answer:

C) 64lb

Explanation:

use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

Vince Weigh in eart  ⇒  X

our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

140pounds    ⇒   56lb

160pounds    ⇒ X

X=\frac{(160)(56)}{140} =64lb

Vince weigh on the other planet is C) 64lb

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3 years ago
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Is this equation balanced? 2HCl + CaCO3 → CaCl2 + H2O + CO2
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3 years ago
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
4 years ago
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