you push a ball to star it rolling along a "perfectly frictionless" surface. How far will the ball roll

A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t

Shalnov [3]

Answer:

a) yield strength

$\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa$

b) modulus of elasticity

strain calculation

$\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046$

strain for offset yield point

$\varepsilon_{new} = \varepsilon_0 -0.002$

=0.0046-0.002 = 0.0026

now, modulus of elasticity

$E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}$

= 184615.28 MPa = 184.615 GPa

c) tensile strength

$\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa$

d) percentage elongation

$\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%$

e) percentage of area reduction

$\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%$

7 0

3 years ago

A man on a bicycle of total mass of 10kg has a Velocity of 2 ms.' He Paddles faster for 5 seconds and the velocity Increase to i

sesenic [268]

Answer:

the value of force, F=4.0N

Explanation:

Firstly, recall velocity-time equation

v=u+at
(4)=(2)+a(5)
a=0.4m/s²

Secondly, recall the Newton's 2nd Law

F=ma
F=(10)(0.4)
F=4.0N

7 0

1 year ago

Light moves at a speed of around 1 million miles per hour O True False

AURORKA [14]

Answer:

False

Explanation:

In miles per hour, light speed is about 670,616,629 mph

5 0

3 years ago

Read 2 more answers

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.