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3 years ago

Match the words with their definitions.

2 answers:

4 0

a crack across rock layers caused by moving plates
molten rock - fault (a fault is a rupture or scission in a block of rocks that allows both parts from the scission to slide over each other)

horizontal intrusion - sill (a sill is a mass of igneous rock that intruded horizontally and laterally through rock layers already existent)

vertical intrusion - d(y)ke (a d(y)ke is also a mass of igneous rock, but this time that intruded vertically across rock layers, many times through an already existing crack)

natural, upward movement of rock layers - uplift (uplift is a geological process in which a portion of the Earth's crust is elevated from its original position due to plate tectonics)

Gennadij [26K]3 years ago

4 0

the first one on his answer is wroung

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: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

Rzqust [24]

Answer:

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

= 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

Therefore the work of the 1.55N 3.36J

4 0

3 years ago

In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separatio

Keith_Richards [23]

Answer:

b. 0.25cm

Explanation:

You can solve this question by using the formula for the position of the fringes:

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

m: order of the fringes

lambda: wavelength 500nm

D: distance to the screen 5 m

d: separation of the slits 1mm=1*10^{-3}m

With the formula you can calculate the separation of two adjacent slits:

$\Delta y=\frac{(m+1)(\lambda D)}{d}-\frac{m\lambda D }{d}=\frac{\lambda D}{d}\\\\\Delta y=\frac{(500*10^{-9}nm)(5m)}{1*10^{-3}m}=2.5*10^{-3}m=0.25cm$