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3 years ago

Which type of stress causes fault-block mountains?

2 answers:

6 0

The kind of stress that cause fault block mountains is tension what are mountains that are produced from tensions stress calls? They called fault block mountains

lakkis [162]3 years ago

6 0

Answer:

tension

Explanation:

correct on edge

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What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

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3 years ago

If an object measures to be 80 centimeters in length, then it is equivalent to ____

Lana71 [14]

Answer:

31.496 inches

Explanation:

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4 years ago

The property of matter that resists changes in motion is called

Nadusha1986 [10]

intertia hope that helps

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4 years ago

Straight wire of indefinite length (transient) passed by an electric current of 5.0 A. The magnetic field generated around this

olchik [2.2K]

Answer:

C. 2.0 cm

Explanation:

The magnetic field around the wire at point M is given by Biot-Savart Law:

$B = \frac{\mu_o I}{2\pi R}$

where,

B = Magnetic field = 50 μT = 5 x 10⁻⁵ T

I = current = 5 A

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

R = distance of point M from wire = ?

Therefore,

$5\ x\ 10^{-5}\ T = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(5\ A)}{2\pi R}\\\\R = \frac{(2\ x\ 10^{-7}\ N/A^2)(5\ A)}{5\ x\ 10^{-5}\ T}\\$

R = 0.02 m = 2 cm

Hence, the correct option is:

7 0

3 years ago

A 0.50 mm-wide slit is illuminated by light of wavelength 500 nm.What is the width of the central maximum on a screen 2.0m behin

Novay_Z [31]

Answer:

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

$y=\frac{n\lambda D}{d}$

where

$\lambda$ is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

$w=2\frac{\lambda D}{d}$

where we have

$\lambda=500 nm = 5\cdot 10^{-7}m$ is the wavelength

D = 2.0 m is the distance of the screen

$d=0.50 mm=5\cdot 10^{-4}m$ is the width of the slit

Substituting, we find

$w=2\frac{(5\cdot 10^{-7} m)(2.0 m)}{5\cdot 10^{-4} m}=0.004 m$

5 0

3 years ago