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Serjik [45]
3 years ago
13

Tectonic plates are pieces of the ________ that float on the more fluid ________ below.

Physics
1 answer:
Alborosie3 years ago
5 0

Answer:

lithosphere and asthenosphere

Explanation:

Lithosphere is the part of the earth surface which is rigid and hard. It consists of whole crust and the upper mantle.It comprises of number of plates called plate tectonic.

The asthenosphere is the Earth's upper mantle's extremely viscous, mechanically fragile and ductile area. This sits below the lithosphere, some 80 to 200 km below the ground at depths. The boundary between lithosphere and asthenosphere is commonly referred to as LAB.

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If you increase the frequency what happens to the distance between the waves (wavelength)?
Dahasolnce [82]

Answer:

C)the distance decreases Shorter wavelength

Step by step Explanation:

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3 years ago
An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass
vladimir2022 [97]
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N 
the slow masses that must be quicker are the pulley, ring, and the rolling sphere. 
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m 
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m 
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
6 0
3 years ago
The diagrams show objects’ gravitational pull toward each other. 3 diagrams labeled X, Y, and Z. X: 2 circles approximatley 2 in
Dafna11 [192]

Answer:C

Explanation:

6 0
4 years ago
Read 2 more answers
Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
4 years ago
Charles Gas Law problem: the temperature inside my refrigerator is about 4 degrees Celsius. If I place a balloon in my fridge th
Harrizon [31]
T1 = 273 + 22 = 295 Kelvin

T2 = 273 + 4 = 277 Kelvin

V1 = 0.5 liters, 


now ,  we just need to input this to the vormula

295/0.5  =  277/V2

590  = 277/V2

V2 = 590/277
     =  2.129 Liter

hope this helps

7 0
3 years ago
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