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3 years ago

Which of the following were utilized in romanesque architecture?

Physics

2 answers:

PolarNik [594]3 years ago

6 0

Towers were used in Romanesque Architecture

LiRa [457]3 years ago

5 0

Hey there,
The answer is D, All of the above
Arches, Vaults and Towers were utilized in Romanesque architecture
Hope this helps :))

~Top

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A rocket will move upward as long as which condition applies?

Dennis_Churaev [7]

The force of thrust is greater than the force if gravity !
Answer found on quizlet !

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3 years ago

For this trajectory, what would the vertical component of acceleration for the module be at time tm=t0−σ=325s? Recall that accel

Karo-lina-s [1.5K]

I think its half of pie so you divid that by 3.14 .

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3 years ago

Read 2 more answers

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

When sedimentary rock is exposed to heat and pressure, what does it change into?

patriot [66]

Metamorphic rock this possess often occurs in the mantle

7 0

3 years ago

Water has the ability to dissolve polar covalent molecules like sucrose by

Alla [95]

Explanation:

Sucrose is a disaccharide which is composed of fructose and glucose. Sucrose molecule has oxygen atoms bonded to hydrogen atoms (O-H bonds - Polar groups) on all ends of its double 6-Carbon ring. The areas near the oxygen atoms are slightly negative, and the areas near the hydrogen atoms are slightly positive that is, the O-H bonds are polar. They bond with the neighbouring Oxygen and Hydrogen atoms because of their

dipole - dipole attractions and hence hydrogen bonds are formed.

However, the covalent bonds within the molecule aren't broken. But rather, the hydrogen bonds holding the sucrose molecules in the crystalline lattice.

5 0

2 years ago