Ask question

Ask question

All categories

3 years ago

7

Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel i

s y=2.0×1011n/m2. how far (δl) would such a string stretch under a tension of 1500 n? express your answer in millimeters using two significant figures.

1 answer:

laiz [17]3 years ago

4 0

L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ = the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer: 15 mm

You might be interested in

if a student lifts their weight of 450 newtons up a set of stairs 5 meters high, how much work did they do?

irina1246 [14]

Answer:

2,250J

Explanation:

W = Fs = (450)(5) = 2,250

4 0

2 years ago

Read 2 more answers

When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration d

Mice21 [21]

The gravitational acceleration at any distance r is given by

$g= \frac{GM}{r^2}$

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is $r_e=6.37 \cdot 10^6 m$ , so the meteoroid is located at a distance of:

$r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m$

And by substituting this value into the previous formula, we can find the value of g at that altitude:

$g= \frac{GM}{r^2} = \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2$

5 0

3 years ago

Read 2 more answers

A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram

Sauron [17]

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

$\rm F=qvBsin\theta$

where

The magnitude of the charge $q=5.7\ \mu C =5.7\times 10^{-6} \ C$

The velocity of the charge $v=4.5\times 10^5\ \frac{m}{s}$

The magnitude of the magnetic field $3.2\ mT=0.0032\ T$

The angle between the directions of v and B $\theta =90^o-37^o=53^o$

By substituting the values we will get:

$F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)$

$F=6.6\times 10^{-3}\ N$

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

To know more about Magnetic force follow

brainly.com/question/14411049

3 0

2 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

3 years ago

Why do the graphs differ?

melisa1 [442]

Well first graph represents rectangular hyperbola

vu = c^2 ( c is constant)

AS 1/v + 1/u = 1/f

Take1/ f to be constant c

1/v = c - 1/u

it is of the form y = - x + k

Slope = -1 having intercept k as shown in fig 2

3 0

3 years ago