1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Zarrin [17]
2 years ago
13

Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin

ation is subjected to a tensile force with magnitude 4000 N.
A) Forthe steel rod, what is the strain?
B) Forthe steel rod, what is the elongation?
C) For the copper rod, what is thestrain?
Physics
1 answer:
nikitadnepr [17]2 years ago
4 0

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

You might be interested in
To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that
noname [10]

Answer:

a. 2v₀/a   b. 2v₀/a  

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. =  1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a  

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s =  v₀(2v₀/a)

= 2v₀/a  

4 0
3 years ago
Please hurry and please no links. I need an answer. Explain why a smart meter monitors the amount of electrical energy being use
brilliants [131]

Smart Metering is a technology encouraging end-use energy efficiency. Smart Meters can give consumers clear and comprehensive information about their energy consumption and, giving them better information, can help consumers to become more energy efficient.

Hope it helped you... pls mark brainliest

4 0
2 years ago
A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i
Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 \frac{m}{s}

Explanation:

The local ocean current is  = 1.52 m/s

Direction \theta = 40°

Velocity component in X - direction V_{x} = 1.52 \cos 40°

V_{x} = 1.164 \frac{m}{s}

Velocity component in Y - direction V_{y} = 8 + 1.52 \sin 40°

V_{y} = 8.97 \frac{m}{s}

The velocity of the ship relative to the earth

V = \sqrt{V_{x}^{2} + V_{y} ^{2}  }

Put the values of V_{x} and V_{y} we get,

⇒ V = \sqrt{1.164^{2} + 8.97 ^{2}  }

⇒ V = 9.05 \frac{m}{s}

This is the velocity of the ship relative to the earth.

7 0
3 years ago
Need help asap please and thank you​
nordsb [41]
Distance = speed x time

distance = 116 x 10

distance = 1160 m
6 0
3 years ago
A 4350 kg truck, driving 7.39 m/s, runs into the back of a stationary car. After the collision, the truck moves 4.55 m/s and the
DiKsa [7]

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

      19792.5+11.5m = 32416.5

        m = 1097.97 kg

Mass of car = 1098 kg

4 0
3 years ago
Other questions:
  • Is walking just a constant state of falling?
    12·1 answer
  • If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa
    5·1 answer
  • A well hole having a diameter of 5 cm is to be cut into the earth to a depth of 75 m. Determine the total work (in joules) requi
    9·1 answer
  • Which of the following statements is true regarding the use of refrigerants?
    11·2 answers
  • Please help please help
    12·1 answer
  • The density of a fish tank is 0.2fish over feet cubed. There are 8 fish in the tank. What is the volume of the tank?
    6·1 answer
  • During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient
    7·1 answer
  • Material speed of light
    10·1 answer
  • A banana has a mass of 1.4 kg. What is its weight, in Newtons?
    14·2 answers
  • How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!