Ask question

Ask question

All categories

2 years ago

8

A force of 5000 n is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a young's modulus of 125 x 1

08 n/m2. the elongation of the rod is:

2 answers:

Blizzard [7]2 years ago

4 0

Por definicion tenemos que
(F/A) = E(∆/0)
Sustituyendo los valores tenemos y despejando ∆:
∆ = (F/(πr2 × E))*0
(5000×5)/(3.14×(34×10^−2)^2×(125×10^8))
5.5×10^−6 m

vladimir2022 [97]2 years ago

4 0

Answer:

The elongation will be 5.51 × 10^-4 m.

Explanation:

Force = F = 5000 N

Length = l = 5.0 m

Area = A = πr^2 = (3.14)(34 × 10^-3)^2 = 3.63 × 10^-3 m^2

Young’s Modulus = E = 125 × 10^8 N/m2

We know that:

E = (F/A)/(∆l/l)

∆l = (F l)/AE

∆l=(5000 ×5)/(3.63 × 10^(-3))( 125 × 10^8)

∆l = 5.51 × 10^-4 m

You might be interested in

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The

Eddi Din [679]

Answer:

(a) Steel rod: $1.1 * 10^{-4}$

Copper rod: $1.88 * 10^{-4}$

(b) Steel rod: $8.3 * 10^{-5} m$

Copper rod: $1.41 * 10^{-4} m$

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = $\frac{1}{Y} * \frac{F}{A}$

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y = 200 000 000 000 $N/m^{2}$

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

∴ $Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}$

For the copper rod:

Y = 120 000 000 000 N/m²

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

$Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain = = 1.88 * 10^{-4}$

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = $1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m$

For the copper rod:

Elongation = $1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m$

6 0

2 years ago

Nancy is pushing her empty grocery cart at a rate of 1.8 m/s. 30 seconds later,

Strike441 [17]

3.Es tarde y mi taxi no llega. Estoy ____.
(5 Points)
preocupada
contenta

3 0

2 years ago

un serbatoio d'acqua di forma cilindrica è riempito con acqua dolce fino a un livello di 6,40 m calcola la pressione dell'acqua

ohaa [14]

Per calcolare la pressione a una certa profondità, devi considerare la legge di Stevino:
<span>p = ρ · g · h

Tenendo conto che:
g = 9,81 m/s²
ρ = 1000 kg/m³

Troviamo:
p(h</span>₁) = ρ · g · h₁ = 1000 · 9,81 · 4,50 = 44145 Pa
p(h₂) = ρ · g · h₂ = 1000 · 9,81 · 5,50 = 53955 Pa

6 0

2 years ago

This sphere is now connected by a long, thin conducting wire to another sphere of radius R2 that is several meters from the firs

alekssr [168]

Here is the full question

A metal sphere with Radius R₁ has a charge Q₁. Take the electric potential to be zero at an infinite distance from the sphere

a) What are the electric field and electric potential at the surface of the sphere?

This sphere is now connected by a long, thin conducting wire to another sphere of radius R₂ that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached:

b) what is the total charge on each sphere?

Assume that the amount of charge on the wire is much less than the charge on each sphere.

Answer:

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b)

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

Explanation:

Given that;

the radius of the sphere = R

The radius of the first sphere = $R_1$

The radius of the second sphere = $R_2$

Charge on the first sphere = $Q_1$

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b) From the question. before the part b question; we learnt that the first sphere is now connected to another sphere;

Now that the two sphere are joined . Charges flows from one to another until their potentials are equal.

As Such; We use $q_1 \ and \ q_2$ to represent their charges respectively

The potential on the surface of the first sphere;

$V_1 = \frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}$

The potential on the surface of the second sphere;

$V_2 = \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

$V_1=V_2$

∴

$\frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}= \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

Thus, we can say :

$\frac{q_1}{q_2}= \frac{R_1}{R_2}$

and $Q_1 = q_1 + q_2$

As such ;

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

3 0

2 years ago

Students are studying the two-dimensional motion of objects as they move through the air. Specifically, they are examining the b

kupik [55]

Answer:

y = - (½ g / v₀²) x²

Explanation:

This is a projectile launch exercise where there is no acceleration on the x-axis so

x = v₀ₓ t

v₀ₓ = v₀ cos tea

y = $v_{oy}$ t - ½ g t2

v_{oy} = v₀ sin θ

as the sphere is thrown horizontally, the angle is tea = 0º, so the initial velocity remains

v₀ₓ = v₀

v_{oy} = 0

we substitute in our equations

x = v₀ t

y = - ½ g t²

we eliminate the time from these equations, we substitute the first in the second

y = - ½ g (x / v₀)²

y = - (½ g / v₀²) x²

this is the equation of a parabola

7 0

2 years ago