The lower mantle being made of solid rock

In which of the following scenarios is the total momentum of the system conserved?

leonid [27]

Answer:

The total momentum of a system is conserved only when the system is closed.

Explanation:

7 0

2 years ago

A certain piece of metal (density 9.45 grams per cubic centimeter) has the shape of a hockey puck with a diameter of 13 cm and a

atroni [7]

Answer:

0.0195 m

Explanation:

$\rho _{p}$ = density of hockey puck = 9.45 gcm⁻³ = 9450 kgm³

$d_{p}$ = diameter of hockey puck = 13 cm = 0.13 m

$h_{p}$ = height of hockey puck = 2.8 cm = 0.028 m

$\rho _{m}$ = density of mercury = 13.6 gcm⁻³ = 13600 kgm³

$d$ = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

$\rho _{m} (0.25)(\pi d_{p}^{2} d ) g = \rho _{p} (0.25)(\pi d_{p}^{2} h_{p} ) g\\\rho _{m} ( d ) = \rho _{p} ( h_{p} )\\(13600) d = (9450) (0.028)\\d = 0.0195 m$

7 0

3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago

What is the weight of a 2.06 kg text book. show work please

salantis [7]

The weight of anything is (mass) x (gravity in the place where the thing is).

 Gravity on Earth is about 9.8 newtons per kilogram, so on Earth,
 the weight of the book is

(2.06 kg) x (9.8 N/kg) = 20.19 newtons .

Another way to get around it is:

 On Earth, 1 kg of mass weighs about 2.205 pounds, so on earth,
 the weight of the book is

 (2.06 kg) x (2.205 pounds/kg) = 4.54 pounds .

7 0

4 years ago

What is an organism's specific role in an ecosystem

Harman [31]

Niche - An organism's particular role in an ecosystem, or how it makes its living.

7 0

3 years ago

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