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3 years ago

PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE

Physics

1 answer:

luda_lava [24]3 years ago

8 0

Australia separated from other continents and species there evolved independently

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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?

jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

diameter of the steel, d = 4 mm

the radius of the wire, r = 2mm = 0.002 m

original length of the wire, L₁

final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)

extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁

the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

$A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2$

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}$

$F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N$

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0

2 years ago

If the velocity of an object is doubled, its kinetic energy is ______

swat32

Increased by a factor of 4

4 0

3 years ago

Suppose an object in freefall is dropped from a building it’s starting velocity is 0m/s. Ignoring the facts of air resistance wh

Alja [10]

v = v₀ + at

v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time

Given values:

v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s

Plug in and solve for v:

v = 0 + 9.81(3)

v = 29.4m/s

3 0

3 years ago

Argue as to why drivint on icy or snowy roads can be dangerous

Vika [28.1K]

Icy/Snowy roads have less friction than normal roads. This means that the wheels are less likely to stay positioned because of traction, and you will spin out of control

6 0

3 years ago

A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete

lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

= 35°

(a)

If friction neglected,

⇒ $F_x=mgSin \theta=ma$

⇒ $a_x=gSin \theta$

$=9.8 \ Sin35^{\circ}$

$=5.62 \ m/s^2$

(b)

If friction present,

⇒ $F_x=mgCos \theta$

⇒ $F_x=mgSin \theta-\mu_r mgCos \theta$

⇒ $a=gSin \theta-\mu_rgCos \theta$

$=9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}$

$=5.46 \ m/s^2$

8 0

3 years ago