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AysviL [449]
3 years ago
12

Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven

much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Physics
1 answer:
faltersainse [42]3 years ago
4 0

Answer:

26 km

Explanation:

Let's say our "cable" has a cross section of 1 m²

Then each meter of cable would weight 7900(9.8) = 77420 N

A Pascal is a Newton per square meter

2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles

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A Cat is on a balcony floor (90cm below the railing), keenly eyeing a butterfly hovering 60 cm above the railing. With what spee
GenaCL600 [577]

We will have the following:

First, the equation to use is the following:

d=v_ot+\frac{1}{2}at^2

Now, we transform the total distance the cat would need to travel:

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So, the cat would need to travel 1.5 meters. ("d" in the equation).

Now, using the speed given we determine the time it would take the cat to traverse the 1.5 meters:

t=\frac{1.5m\cdot1s}{0.45m}\Rightarrow t=\frac{10}{3}\Rightarrow t=3.333\ldots

So, the time it would take the cat to traverse the distance will be approximately 3.33 seconds.

Now, we know that the acceleration will be given by Earth's gravity, so:

1.5m=v_0(\frac{10}{3}s)+\frac{1}{2}(-\frac{9.8m}{s^2})(\frac{10}{3}s)^2\Rightarrow1.5m=v_0(\frac{10}{3}s)+(-\frac{490}{9}m)\Rightarrow\frac{1007}{18}m=v_0(\frac{10}{3}s)\Rightarrow v_0=\frac{1007}{60}\frac{m}{s}\Rightarrow v_0=16.78333\ldots

So, the initial vvelocity the cat must leave the floor in order to arrive at the butterfly with the optimum pouncing speed of 0.45 m/s is approximately 16.78 m/s or exactly 1007/60 m/s.

5 0
2 years ago
A 6.00 kg box is resting on a table. What is the magnitude of the normal force of the table on the box?
Amiraneli [1.4K]

Answer:

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3 years ago
100 points!!!! Will mark brainliest!!!!!!!!
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Answer:

Normal Force, Applied force, Gravitational Force, Spring Force

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Explained in the attachment !

<h3>Hope it helps you!!</h3>

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