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2 years ago

What stress causes this type of fault to form?

Physics

2 answers:

Mars2501 [29]2 years ago

4 0

Answer:

The answer to this question is compression.

worty [1.4K]2 years ago

3 0

Answer:

gravity

Explanation:

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For a snowboard jumper in the air, what force or forces will be most important for modeling the motion?

Reptile [31]

Answer: Gravitational force and drag force

Explanation:

For a snowboard jumper in the air, two forces would be acting. One in the downward direction- the gravitational pull and second in the opposite direction to the motion, the drag force due to air. If the snowboard jumper jumps in the air at a certain angle with the horizontal. The forces are written as the sum of vertical and horizontal components. Hence, for the modeling the motion, gravitational force and drag force are important,

4 0

2 years ago

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

1 year ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

2 years ago

The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving ele

siniylev [52]

The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving electrons form a electron negative blanket that binds the atomic positive nuclei together, forming a metallic bond.

So the answers are { Negative } and { Positive }.

Please vote Brainliest (:

5 0

2 years ago

Read 2 more answers

Which of these is not a type of force?

coldgirl [10]

Inertia
the awnswer is inertia b

6 0

3 years ago

Read 2 more answers