8. Differentiate between each of these land designations made during the Progressive Era:

At 20 C, a steel rod of length 40.000 cm and a brass rod

balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

Step 1: identify the given parameters and standard parameters

⇒Steel rod: length of rod = 40.000 cm

Young modulus(Υ) = 20 X 10¹⁰ pa

coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

stress in the rod =?

⇒Brass rod: length of rod = 30.000 cm

Young modulus(Υ) = 9 X 10¹⁰ pa

coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

stress in the rod =?

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Step 2: calculate the stress in each rod

⇒Steel rod: stress in the rod = -Y*α*ΔT

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

= -9.6 X 10⁷ pa

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⇒Brass rod: stress in the rod = -Y*α*ΔT

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

= -7.2 X 10⁷ pa

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∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

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Step 3: calculate the new length of each rod

⇒Steel rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

New length = ΔL + L₀ = 0.0192cm + 40cm

New length = 40.0192cm

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⇒Brass rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

ΔL = 2400 X 10⁻⁵ cm = 0.024cm

New length = ΔL + L₀ = 0.024cm + 30cm

New length = 30.024cm

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Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

8 0

2 years ago

What is the speed of a car traveling 100 m in 20 s?

Ludmilka [50]

Answer: To answer this question, we will need the following equation: SPEED = DISTANCE/TIME (A multiplication and division triangle will be shown)i) The speed of the car is calculated by doing 100 metres/ 20 seconds which gives us 5 metres per second. ii) Rearranging the equation earlier, we can make the distance the subject of the equation so that we get SPEED x TIME = DISTANCE. We worked out the speed and the time was given as 1 minute 40 seconds but we cannot plug in the numbers yet as the time has to be converted to units of seconds (because our speed is in meters per second). 1 minute 40 seconds = 60 seconds + 40 seconds = 100 secondsWe then plug in the numbers to get the distance travelled = 5 metres per second x 100 seconds = 500 metres.

Explanation:

3 0

2 years ago

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

6 0

2 years ago

How much time would it take for an object to fall 4.7 meters

levacccp [35]

Answer:

4.7 is 10 as much as the number 0.47.

If you multiply 0.47 x 10 it will equal 4.7

Explanation:

4 0

2 years ago

Which statements about a neutral atom are correct? Select all that apply. (1) A neutral atom is composed of both positively and

Bad White [126]

Answer:

(3) The electrons are attracted to the positively charged nucleus.

(4) Positively charged protons are located in the tiny, massive nucleus.

(6) The negatively charged electrons are spread out in a "cloud" around the nucleus.

Explanation:

An atom is the basic particle in which a matter is formed. It can either be charged or neutral. When it losses or gains an extra electron, it becomes charged, an ion. While a neutral atom is an atom which has no charge, because it has the same number of electrons and protons. The protons are located in the tiny massive nucleus of the atom, while the electrons are in orbits or cloud around the nucleus.

7 0

3 years ago