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lord [1]
3 years ago
8

What are the characteristics of Earth’s lithosphere and how does the lithosphere play a role in plate tectonics?

Physics
1 answer:
pashok25 [27]3 years ago
4 0
The lithosphere includes the brittle upper portion of the mantle and the crust, the outermost layers of Earth's structure. It is bounded by the atmosphere above and the asthenosphere (another part of the upper mantle) below. Although the rocks of the lithosphere are still considered elastic, they are not viscous
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I NEED THIS ASAP!!! A ball is thrown straight up with an initial velocity of 4.40 m/s. Assuming there is no air friction, what i
Lilit [14]

Answer:

I think it is 80m/s

Explanation:

d = ½ g t2

  = ½ (10 m/s2) (4 s)2

  = (5 m/s2) (16 s2)

  = 80 ms

75% sure

Hope this helps!!!

5 0
3 years ago
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
Which is not expecting acceleration?
Kamila [148]

1. Answer: A skydiver whose air resistance is equal to that of her weight.

A skydiver free falls under gravity but her rate of fall slows down due to drag -air resistance. when this air resistance becomes equal to her weight, the two get balanced and the body does not accelerate or decelerates.

2. Answer: Gravity

Contact forces are those which act when there is physical contact between two bodies. For example: normal force, tension and spring force.

Non-contact forces act between two bodies even when they are at a distance apart. For example: gravity, electric force, magnetic force etc.

3. Answer: The tendency of an object's motion to remain the same.

Inertia is a property of matter by virtue of which it tends to remain in its state of motion or rest. It does depend on mass of the object, more the mass, more is inertia. For example, cycle can be easily moved but we need real push hard for a car to move.

4. Answer: 254 N

The man pushes the box with 310 N force at an angle of  55 degrees to the horizontal.

we can write this in terms of horizontal (F cos \theta)and vertical component (F sin\theta).

Horizontal component: F_H=310 \times cos (55^o)= 178 N

Vertical component: F_v=310 \times sin (55^o)= 254 N

The vertical component would act towards the floor making the job more difficult to move the job.



7 0
3 years ago
Your high beam headlights illuminate the road in front of you for __________ feet. A. 150 B. 450 C. 650
ASHA 777 [7]

Answer:

B. 450 feet

Explanation:

Due to the angle at which high beam headlights illuminate, they can illuminate the road for about 450 feet.

8 0
3 years ago
Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
5 0
3 years ago
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