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liubo4ka [24]
3 years ago
10

Sophie says that geologic maps do not matter because she gets no benefits from them. Why is Sophie wrong? a. She can see the ran

ges of mountains near her house. B. She needs to learn how to read geologic maps in school. C.She benefits from products that use minerals found using geologic maps. D.She can see the weather and dress appropriately.
Physics
2 answers:
Oliga [24]3 years ago
8 0
A geologic map<span> or </span>geological map<span> is a special-purpose </span>map<span> made to show </span>geological<span> features. 

That being said, C should be the correct answer. Many products that she benefits by have to be found with geologic maps.</span>
astra-53 [7]3 years ago
4 0
The Answer is Option C 
I think...
Sorry If i am wrong...
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A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten
satela [25.4K]

Answer: 20m

Explanation:

We will solve this question by applying the law of conservation of energy which states that the sum of potential energy and kinetic energy is always the same.

The PE is 0 at surface and maximum at top while the KE is maximum at surface and 0 at top.

From the question,

PE = mgh = 50 J -(1)

mg* 10 = 50

mg = 50/10

mg = 5

The total energy at that point = PE + KE = 50 + 50 = 100 J

Therefore, at topmost point, the PE will be 100 J

mgH = 100J , H is the needed height

Using the value of mg obtained above, we have

H= 100/5

H = 20 m

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3 years ago
A 70 kg student is riding on a skateboard at 4m/s. A 60 kg student is also riding a
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What is a function of the pineal gland in the endocrine system
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A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a
stira [4]

Answer:

Explanation:

Given

acceleration is given by

a=-g-c_Dv^2

where \ddot{y}=a

\dot{y}=v

Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ds=\frac{v}{a}dv

\int ds=\int \frac{v}{-g-0.001v^2}dv

\Rightarrow Let -g-0.001v^2=t

-0.001\times 2vdv=dt

vdv=-\frac{dt}{0.002}

at\ v_0=50\ m/s,\ t=-g-0.001(50)^2

t=-g-2.5

at v=0,\ t=-g

\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}

\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})

s=113.608\ m

when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

h=127.55\ m

3 0
3 years ago
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