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3 years ago

What type of fault creates huge mountains

1 answer:

vladimir1956 [14]3 years ago

7 0

Together, normal and reverse faults are called dip-slip faults, because the movement on them occurs along the dip direction -- either down or up, respectively. Reverse faults create some of the world's highest mountain chains, including the Himalaya Mountains and the Rocky Mountains .

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Two energy transfers take place when a book hits the ground, Which type of energy transfers are those

bazaltina [42]

Answer:

I think that when a book hits the ground its potential energy converts into kinetic energy and then kinetic energy is transformed into sound and heat energy.

Explanation:

6 0

3 years ago

What is the equation for an inelastic collision

abruzzese [7]

M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.

7 0

3 years ago

A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3

marissa [1.9K]

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

$mg(H-h)=\frac{1}{2}mv^2$

$g(H-h)=\frac{1}{2}v^2$

$\Rightarrow H=\frac{v^2}{2g}+h$

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

$=v\frac{2g}{h}$

$\Rightarrow v^2=\frac{gD^2}{2h}$

thus, height H in terms of D is given by

$H=\frac{D^2}{2h}+h$

$H=\frac{13.9^2}{2\times3}+3$

H=35.20 m

4 0

3 years ago

Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs down

MA_775_DIABLO [31]

Answer:

a) D_ total = 18.54 m, b) v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

sin 45 = y₁ / d

cos 45 = x₁ / d

y₁ = d sin 45

x₁ = d sin 45

y₁ = 8 sin 45 = 5,657 m

x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

y₂ = 12 m

x₂ = 0

total displacement

y_total = y₁ + y₂

y_total = 5,657 + 12

y_total = 17,657 m

x_total = x₁ + x₂

x_total = 5,657 + 0

x_total = 5,657 m

D_total = 17.657 i^+ 5.657 j^ m

D_total = Ra (17.657 2 + 5.657 2)

D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

v = Δx /Δt

the distance traveled using the pythagorean theorem is

r = √ (d1² + d2²)

r = √ (8² + 12²)

r = 14.42 m

The time used for this shredding is

t = t1 + t2

t = 1 + 1.2

t = 2.2 s

let's calculate the average speed

v = 14.42 / 2.2

v = 6.55 m / s

8 0

3 years ago

Inside a television picture tube there is a build-up of electrons (charge of 1.602 × 10^–19 C) with an average spacing of 38.0 n

Brut [27]

The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9 N. m2 / C2

Given are the following:
Q = 1.602 × 10^–19 C
r = 38 x 10^-9 m

Substitue the given:
E = $\frac{( 1.602 x 10^{-19} )( 9.0x10_{9} )}{(38x10^{-9}) ^{2} }$

E = 998.476 kN/C

8 0

3 years ago

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