Which option correctly defines the scientific method?

A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r

katrin2010 [14]

Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force(F)/Area(A)

While ε_z is longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

6 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

Compare the forces in a small nucleus to the forces in a large nucleus

pochemuha

The comparison of the forces in a small nucleus to the forces of a large one is the fact that they are capable of holding the protons and neutrons which made it no matter what their size may be. Therefore, as long as there is a nucleus, their forces can both hold together the two atoms tight.

6 0

3 years ago

A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stay

yuradex [85]

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = $20\times 1=20\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2$

$v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s$

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

4 0

3 years ago

Two forces are applied to a 17 kg box, as shown. The box is on a smooth surface. Which statement best describes the acceleration

RoseWind [281]

To the picture the answer is A. I can’t answer the typed question because I need the picture for the box

7 0

3 years ago