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alexandr1967 [171]
3 years ago
5

What can scientist learn from studying sedimentary rocks?

Physics
2 answers:
cricket20 [7]3 years ago
7 0
They are made of rocks
Andrej [43]3 years ago
4 0
Scientists can learn by analysing the layers of the rock
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The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t
Bumek [7]

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of 8.1\times 10^5 N

Mass of train=1.9\times 10^7

Final speed (v)=68 km/h \approx 18.88 m/s

F=ma

so acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}

a=0.042631 m/s^2

and acceleration is

a=\frac{v-u}{t}

0.042631=\frac{18.88-0}{t}

t=443.089 \approx 443.1 s

8 0
3 years ago
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a lawn mower is pushed with a force of 50n. if the angle between the handle of the mower and the ground is 30°. why doesn't the
Mademuasel [1]

Answer:

IT doesnt push down because the lawnmower has wheels and the ground is hard

Explanation:

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5 0
3 years ago
How fast did Newton decide a cannon ball shot out in order to escape Earth’s gravity?
arsen [322]

Plugging in for the Earth's mass and for G, we have 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is about 25,000 miles per hour

7 0
3 years ago
A non-_____ rock has interlocking grains with no specific pattern.
Kazeer [188]
A non <span>foliated </span>rock has interlocking grains with no specific pattern.
3 0
3 years ago
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The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc
Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is 1.163 * 10^{-5} m

Explanation:

From the uncertainty Principle,

Δ_{y} Δ_{p} = \frac{h}{2\pi }

The momentum P_{y} = (mass of electron)(speed of electron)

                                = (9.109 * 10^{-31}kg)(995 * 10^{3} m/s)

                                = 9.0638 * 10^{-25}kgm/s

If the uncertainty is reduced to a 0.0010%, then momentum

                              = 9.068 * 10^{-30}kgm/s

Thus the uncertainty in the position would be:

                              Δ_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }

                              Δ_{y} \geq  1.163 * 10^{-5}m

5 0
3 years ago
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