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2 years ago

What observations can the geologist make by working outdoors instead of in a lab?

Physics

1 answer:

vredina [299]2 years ago

8 0

Answer:

Geology is the study of the Earth that involves the process at Earth, materials of which it is made, and its history.

<u>Geologists combine both laboratory and field data to illustrate the results of their research. Some observations that can the geologist make by working outdoors instead of in a lab are as follows:</u>

Understanding and exploring the earth's surface closely using geophysical tools.
Collecting samples by own and make some interpretations at the same time.
Observation of the landscapes
Close observation of outcrops

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If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step

azamat

Answer:

Explanation:

$v_f^2 = v_i^2-2a(x_f-x_i)$

Subtract both sides by $v_i^2$ :

$- v_i^2+v_f^2 = -2a(x_f-x_i)$

Divide both sides by -2*a:

$\frac{v_i^2 - v_f^2}{2a} =x_f-x_i$

Add both sides by $x_i$ :

$x_i+\frac{v_i^2 - v_f^2}{2a} =x_f$

Subtract both sides by $\frac{v_i^2 - v_f^2}{2a}$ :

$x_i=x_f-\frac{v_i^2 - v_f^2}{2a}$

8 0

3 years ago

Which of the following statements is FALSE regarding the law of conservation of energy?

Jobisdone [24]

Answer:

Explanation:

because it doesn't agree with the law of definite proportion

and energy can niether be created nor destroyed but can be transformed from one form to another

5 0

2 years ago

Energy is converted from solar to chemical in Process A and then from one form of chemical to another in Process B. Process A is

Pepsi [2]

I think the answer is photosynthis, when plants turn light into food and energy.

9 0

3 years ago

Read 2 more answers

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in

USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

$W=F\cdot d$

Then, we can derive the magnitude of the force as:

$F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N$

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

$F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N$

6 0

2 years ago