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2 years ago

Using evidence from the article, defend the concept that Earth’s magnetic poles have swapped places over time.

1 answer:

6 0

Answer: Scientists found evidence of Earths magnetic field reversal in rocks on the ocean floor at plate boundaries. These rocks have alternating polarity due to magnetization that occurred during their cooling period. Using radio metric dating, scientist estimate that reversals occur approximately every several hundred thousand years.

Explanation:

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

2 years ago

A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50

kolezko [41]

Answer:

$Distance_{vaccum}=5.19cm$

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

$T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}$

$T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}$

Thus in the same time the it would had covered a distance of

$Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]$

we have

$\mu _{ice}=1.309\\\\\mu _{quartz}=1.542$

Applying values we have

$Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]$

$Distance_{vaccum}=5.19cm$

6 0

3 years ago

What is the force required to accelerate a baseball with a mass of 0.145 kilograms and an acceleration of 35.00 meters/second sq

Paladinen [302]

Answer:5.075N

Explanation:

Mass=0.145kg

Acceleration=35m/s^2

Force=mass x acceleration

Force=0.145 x 35

Force=5.075N

3 0

2 years ago

Solve for the unknown in each of these circuits

Tamiku [17]

Ohms law: This law relates voltage difference between two points. Mathematically, the law states that V=IR;

Where

V = voltage difference ; in volts

I = Current ; in Amperes

R = Resistance ; in ohms

1. Answer : given that R = 10 ; V= 12 V ; I = ?

From ohms law, I = V/R

= 12/10

= 1.2 Amp.

2. Answer: given that R = 10 ; V= ? ; I = 5

From ohms law, V = IR

= 10×5 = 50 V

3 . Answer: given that R = ? ; V= 120 ; I = 5

From ohms law, R = V/I

= 120/5

= 24 Ω

4 . Answer: given that R = ? ; V= 10 ; I = 20

From ohms law, R = V/I

= 10/20

= 0.5 Ω

5 . Answer: given that R = 480 ; V= 24 ; I = ?

From ohms law, I = V/R

= 24/480

= 0.05 A

6. Answer: given that R = 150 ; V= ? ; I = 1

From ohms law, V = IR

= 1 × 150

= 150 V

6 0

3 years ago

Introduction to Forces Warm-Up Active How do forces affect the motion of an object?

In-s [12.5K]

Answer:

Forces can affect an object.

Balanced forces allow an object to continue moving at a constant motion (law of inertia).

Unbalanced forces cause a change in motion.

4 0

1 year ago

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