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Anon25 [30]
3 years ago
9

What was the main view how the world worked geologically prior to the 1960s? It was generally believed that continents and ocean

s moved. It was generally believed that mountains were produced by vertical forces. It was generally believed that oceans formed as a result of meteorite impacts. It was generally believed that the earth was flat.
Physics
1 answer:
attashe74 [19]3 years ago
3 0

Answer:

It was generally believed that mountains were produced by vertical forces

Explanation:

                   The main view of the world worked geologically prior to the 1960s was that the mountains were formed by the vertical forces of nature.

                    The early people prior to 1960s believed in many different natural phenomenons and they give their own reasons for their occurrence. But later many researchers and geophysicists studied the formation of the earth and came with possible answers to these questions.

Thus the answer is  

" It was generally believed that mountains were produced by vertical forces."

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What are the benefits of the cool-down period following exercise?
kvv77 [185]

The benefits of the cool down period are quite important, it allows your body to slow your heart rate at a nice healthy safe pace, if you stop right away it can cause breathing, heart, and muscle problems.


5 0
3 years ago
When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N

Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
3 years ago
A 50-kg copper block initially at 140°c is dropped into an insulated tank that contains 90 l of water at 10°c. Determine the f
xxMikexx [17]

Answer:

T_f=24.71

Explanation:

From the question we are told that:

Mass of block m=50

Temperature of block T_b =140 \textdegree C

Volume of water V= 90L

Temperature of water T_w=10 \textdegree C

Density of water \rho=1000kg/m^3

Specific heat of water C_w=4.18KJ/kg-k

Specific heat of copper C_p=0.96KJ/kg-k

 

Generally the equation for equilibrium stage is mathematically given by

mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)

50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)

48(140-T_f)=376.2(T_f-10)

140-T_f=7.8375(T_f-10)

140-T_f=7.8375T_f-78.375

-8.8375T_f=-218.375

T_f=\frac{-218.375}{-8.8375}

T_f=\frac{-218.375}{-8.8375}

T_f=24.71

6 0
3 years ago
Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm​
mixer [17]
<h2>~<u>Solution</u> :-</h2>
  • Here, the <u>moment arm</u> is defined as follows;

The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.

\\

7 0
3 years ago
On a small planet, an astronaut uses a vertical
sleet_krkn [62]
Constant velocity means the netto force = 0, therefore F(gravity) = F(astronaut).
175N divided by 87,5kg = 2.00kg/N
6 0
3 years ago
Read 2 more answers
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