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horsena [70]
3 years ago
10

What is the final speed of a 60 kg boulder dropped from a 111 meter cliff

Physics
1 answer:
saveliy_v [14]3 years ago
3 0

After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is

       Square root of ( 19.6 M ) .

If M=111 meters, then the speed is <em>46.64 meters per second</em>.

We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.



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andrey2020 [161]
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and
"what are the best shoes to wear when exercising"
Both of these questions can be tested in a scientific way through an experiment. 
5 0
3 years ago
Which factors affect the strength of the electric force between two objects
Margaret [11]
<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>
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3 years ago
Read 2 more answers
Five hundred million years ago, basaltic lava flowed in an area now known as Monticello, the historic home of Thomas Jefferson.
aev [14]

Explanation:

               Igneous -  metamorphic -  sedimentary

A rock cycle provides the cyclic transformation of one rock type to another in nature.

There are three main types of rock involved in the rock cycle;

  • igneous rocks are derived from the cooling and solidification of molten magma
  • metamorphic rocks are changed rocks subjected to intense pressure and temperature
  • sedimentary rocks are derived from rock sediments that have been lithified.

The history of the rock in Monticello begins with igneous rock formation. Basalt is an igneous rock that forms from the cooling and solidification of molten magma. Under intense pressure and temperature regimes, they are changed to metamorphic rocks.

Agents of denudation such as wind, water and glacier weathers the rock and disintegrates it. They are then carried into basins where they are deposited. Here they form sedimentary rock.

The process still goes on as the sedimentary rock gets taken into depth, they can either melt to form igneous rock or be changed to metamorphic rocks.

learn more:

metamorphic process brainly.com/question/869769

sedimentary rocks brainly.com/question/9131992

#learnwithBrainly

5 0
3 years ago
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
A amusement park moves riders in a circle at a rate of 6.0m/s if the radius is 9.0 meters what is the acceleration of the ride
oksano4ka [1.4K]

Centripetal acceleration is (speed-squared) / (radius)

CA = (6 m/s)² / (9 m)

CA = (36 m²/s²) / (9 m)

CA = (36/9) (m²/m·s²)

<em>Centripetal acceleration = 4 m/s²</em>

4 0
3 years ago
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