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Tasya [4]
3 years ago
12

What type of subsurface material is able to store groundwater?

Physics
1 answer:
notka56 [123]3 years ago
4 0
<span>Porous material has many spaces that can hold(store) groundwater.</span>
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A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
4 years ago
Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom
Irina-Kira [14]

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt  where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i  -  .46 x 1.9³ / 3 j

= .514i - 1.05 j

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

x component = - 2.586

y component = 2.85

7 0
4 years ago
what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I
denis23 [38]

Answer: hello some part of your question is missing attached below is the missing detail

answer :

<em>w</em>f = M( v cos∅ )D / I

Explanation:

The Angular speed <em>wf  </em>of the system after collision in terms of the system parameters  and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3  + mD^2 ) <em>w</em>f

where ;  ( ML^2 / 3  + mD^2 ) = I ( Inertia )

In terms of system parameters and I

<em>w</em>f = M( v cos∅ )D / I

5 0
3 years ago
How to Convert N/m2 to N/cm2 ?​
Brut [27]

1 newton / square metre =0.0001 newton / square centimetre

Formula = divide the pressure value by 10000

5 0
3 years ago
6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory
Svetllana [295]

Answer:

Engular velocity: w=1,24*10^{-3}/s

Linear velocity: V=7905 m/s

The time it takes:

t=5060s=84min

Explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1)a=r*w^{2}

Solving for w:

(2)w=\sqrt{\frac{a}{r} }

Replacing a=9,8m/s2 and r=6,375,000m:

(3)w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s

And the angular velocity relates to the linear velocity:

V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s

The perimeter of the orbit is:

P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m

The time it takes:

t=\frac{P}{V} =\frac{40.05*10^{6}m}{7905 m/s}=5060s=84min

5 0
3 years ago
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