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Helen [10]
2 years ago
15

Descirbe what has happen to earths continents today

Physics
1 answer:
kakasveta [241]2 years ago
5 0
Today, we know that the continents rest on massive slabs of rock called tectonic plates. The plates are always moving and interacting in a process called plate tectonics. The continents are still moving today. Some of the most dynamic sites of tectonic activity are seafloor spreading zones and giant rift valleys.
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Help pleaseeeeeeeeeeeee
Tasya [4]
A it’s not that hard tbh
7 0
3 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
3 years ago
The gravitational force between two objects depends on the masses and what factor between them?
Charra [1.4K]
It depends on mass and distance.  
8 0
3 years ago
A cube has a density of 1900 kg/m 3 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experime
Deffense [45]

Answer:

4847.94844926 kg/m³

Explanation:

\rho' = Actual density of cube = 1900 kg/m³

\rho = Density change due to motion

v = Velocity of cube = 0.92c

c = Speed of light = 3\times 10^8\ m/s

Relativistic density is given by

\rho=\dfrac{\rho'}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-\dfrac{0.92^2c^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-0.92^2}}\\\Rightarrow \rho=4847.94844926\ kg/m^3

The cube's density as measured by an experimenter in the laboratory is 4847.94844926 kg/m³

3 0
3 years ago
A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ
3241004551 [841]

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

PE_i + KE_i = PE_f + KE_f

mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2

now here we know that

h_1 = 2.75 m

v_i = 0

v_f = 5.23 m/s

now plug in all values in above equation

mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2

divide whole equation by mass "m"

9.8*2.75 = 9.8*h + \frac{1}{2}*27.35

9.8*h = 13.27

h = 1.35 m

so height of the ball from ground will be 1.35 m

4 0
3 years ago
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