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iren [92.7K]
3 years ago
8

PLEASE HELP!!

Physics
1 answer:
Klio2033 [76]3 years ago
6 0

Answer:

28.15

Explanation:

the ateps is in the photo

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Which statement is not a good practice when working inside a computer case?
lozanna [386]

Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

3 0
3 years ago
after a large snowstorm you shovel 2000 kilograms of snow off your side walk in 1 hour. you lift the shovel to an average height
solmaris [256]

Lifting a mass to a height, you give it gravitational potential energy of

       (mass) x (gravity) x (height)  joules.

To give it that much energy, that's how much work you do on it.

If 2,000 kg gets lifted to 1.25 meters off the ground, its potential energy is

         (2,000) x (9.8) x (1.25) = 24,500 joules.

If you do it in 1 hour (3,600 seconds), then the average power is

           (24,500 joules) / (3,600 seconds) = 6.8 watts.

None of these figures depends on whether the load gets lifted all at once,
or one shovel at a time, or one flake at a time.

But this certainly is NOT all the work you do.  When you get a shovelful
of snow 1.25 meters off the ground, you don't drop it and walk away, and
it doesn't just float there. You typically toss it, away from where it was laying
and over onto a pile in a place where you don't care if there's a pile of snow
there. In order to toss it, you give it some kinetic energy, so that it'll continue
to sail over to the pile when it leaves the shovel.  All of that kinetic energy
must also come from work that you do ... nobody else is going to take it
from you and toss it onto the pile.


8 0
3 years ago
The elements are mostly likely to form more than one type of ion are the
Ghella [55]

Alkali Metals ......................................

8 0
3 years ago
Extra water is absorbed from undigested food in the _____.
Nataly_w [17]

Answer: large intestine

Explanation:

mouth grind and moisten food

saliva break down starch into sugar

esophagus transport food from mouth to stomach

stomach churn food, break down food with enzymes

liver add bile to break down fat

pancreas add enzymes to digest protein

gallbladder store bile

small intestine absorb food through villi

large intestine absorb extra water and remove waste from body

5 0
3 years ago
The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

h_{Hg-TOP}=675mmHg=0.675m the barometric reading at the top of the building

h_{Hg-BOT}=695mmHg=0.695m the barometric reading at the bottom of the building

\rho _{air}=1.18 kg/m^{3} density of air

\rho _{Hg}=13600 kg/m^{3} density of mercury

g=9.8/m^{2} gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building P_{TOP} and the perssure at the bottom P_{BOT}:

P_{TOP}=\rho _{Hg} g h_{Hg-TOP} (1)

P_{BOT}=\rho _{Hg} g h_{Hg-BOT} (2)

From (1): P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa (3)

From (2): P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure \Delta P:

\Delta P=P_{BOT} - P_{TOP} (5)

\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa (6)

On the other hand, we have a column of air with a cross-section area A and the same height of the building, lets name it h_{air}.

As pressure is defined as the force F exerted on a specific area A, we can write:

\Delta P=\frac{F}{A} (7)

If we isolate F we have:

F= A \Delta P (8)

Also, the force gravity exerts on this column of air (its weight) is:

F=m_{air} g (9)

Knowing the density of air is: \rho_{air}=\frac{m_{air}}{V_{air}} (10)

where the volume of air can be written as: V_{air}=(A)(h_{air}) (11)

Substituting (1) in (10):

\rho_{air}=\frac{m_{air}}{(A)(h_{air}} (12)

Isolating m_{air}:

m_{air}=(\rho_{air}) (A) (h_{air}) (13)

Substituting (13) in (9):

F=(\rho_{air}) (A) (h_{air}) (g) (14)

Matching (8) and (14)

A \Delta P=(\rho_{air}) (A) (h_{air}) (g) (15)

Isolating h_{air}:

h_{air}=\frac{\Delta P}{g \rho_{air}} (16)

Substituting the known and calculated values:

h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})} (17)

Finally:

h_{air}=230.50 m This is the height of the building

8 0
3 years ago
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