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3 years ago

Which compares the thermal energy found in a piece of coal with the thermal energy found in a newly opened coal mine?

1 answer:

4 0

<span>The thermal energy found in a piece of coal is lesser than in the thermal energy in a newly opened coal mine. The newly opened coal mine requires a greater thermal energy to really open it due to its solid structure. Whereas in a piece of coal, it only requires a lesser amount of thermal energy to extract it.</span>

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When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic

USPshnik [31]

Total energy is a spring:
$E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2$

At x = 0.5a:
$\frac{1}{2} k \frac{a}{2} ^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2 \\ \frac{1}{2} mv^2 = \frac{1}{2} ka^2 - \frac{1}{8} ka^2 = \frac{3}{8} ka^2$

The ration:
$\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} = \frac{3}{4}$

5 0

3 years ago

Read 2 more answers

Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

8 0

2 years ago

The density of aluminum is 2.7 g/cm3. A metal sample has a mass of 52.0 grams and a volume of 17.1 cubic centimeters. Could the

Fudgin [204]

Answer:
__________________________________________________
No; the sample could not be aluminum;
since the density of aluminum, " 2.7 g/cm³ " , is NOT close enough to the density of the sample, " 3.04 g/cm³ " .
________________________________________________
Explanation:
________________________________________________
Density is expressed as "mass per unit volume" ;

in which:
"mass, "m", is expressed in units of "g" (grams); and:
"Volume, "V", is expressed in units of "cm³ " (such as in this problem); or in units of "mL" ;
__________________________________________________
{Note the exact conversion: " 1 cm³ = 1 mL " .}.
__________________________________________________
The formula for density: D = m/V ;

Given: The density of aluminum is: 2.7 g/cm³.

Given: A sample has a mass of 52.0 g ; and Volume of 17.1 cm³ ; could it be aluminum?
_________________________________________________________
Let us divide the mass of the sample by the volume of the sample;
by using the formula:
___________________________________________
D = m / V ;

and see if the value is at, or very close to "2.7 g/cm³ ".

If it is, then it could be aluminum.
____________________________________________________
The density for the sample:

D = (52.0 / 17.1) g/cm³ = 3.0409356725146199 g/cm³ ;
→round to "3 significant figures" ;
= 3.04 g/cm³ .
_______________________________________________
No; the sample could not be aluminum; since the density of aluminum,
"2.7 g/cm³ " is NOT close enough to the density of the sample,
"3.04 g/cm³ " .
____________________________________________________

5 0

3 years ago

How did Galileo increase public support for Copernicus’s model?

miskamm [114]

I think by using data collected by Tycho Brahe

8 0

2 years ago

Read 2 more answers

The position-time equation for a cheetah chasing an antelope is:

pochemuha

Answer:

x = 1.6 + 1.7 t^2 omitting signs

a) at t = 0 x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0 V = 0

c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)

d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)

4 0

2 years ago