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Natasha2012 [34]
3 years ago
12

As plates move, large blocks of crust ___ along the faults

Physics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

move

Explanation:

i think that is it

as plates move large blocks of crust move along the faults

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Is lateral shift or lateral displacement same ?
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When a ray of light passes through a glass slab of a certain thickness, the ray gets displaced or shifted from the original path. This is called lateral shift/displacement.

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What is a compound and a element ​
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A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by u = (5t 2) rad
OverLord2011 [107]

Answer:

Explanation:

Given

radius of  path r=2\ m

Velocity of Particle \theta =5t^2 rad

where t=time in seconds

angular velocity of particle is given by

\omega =\frac{\mathrm{d} \theta }{\mathrm{d} t}

\omega =2\times 5t=10\cdot t

And angular acceleration is given by

\alpha =\frac{\mathrm{d} \omega }{\mathrm{d} t}

\alpha =10 rad/s^2

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a_t=10\times 2=20\ m/s^2

Centripetal acceleration a_c=\omega ^2\times r

a_c=(10t)^2\times 2=200t^2

net acceleration is sum of tangential and centripetal force at any time t is given by

a_{net}=\sqrt{(a_c)^2+(a_t)^2}

a_{net}=\sqrt{(200t)^2+(20)^2}

a_{net}=20\sqrt{(10t)^2+1}\ m/s

                 

8 0
3 years ago
a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while th
anzhelika [568]

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

F_{1} = 250

A_{1} = Area of the small piston = πr^{2}

A_{1} = 22/7 x 0.4^{2}

A_{1} = 0.5 m^{2}

F_{2} = ?

A_{2} = Area of the large piston = πr^{2}

A_{2} = π x 1

A_{2} = 3.14 m^{2}

To calculate the force on the large piston, we will use the below formula

F_{1}/ A_{1} = F_{2} / A_{2}

Substitute all the parameters into the equation

250/0.5 =  F_{2}/3.14

F_{2} = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

F_{1}/ 0.25A_{1} = F_{2} / A_{2}h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

Learn more here: brainly.com/question/13596980

5 0
2 years ago
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