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Blababa [14]
3 years ago
6

The pressure of a liquid is given by P = pgh. Calculate the pressure (in SI unit) if the

Physics
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Answer:

5000 Pa

Explanation:

First collect the data you've been given already and make sure to convert into the right units;

<em>Density</em><em> </em><em>=</em><em> </em><em>1</em><em> </em><em>g</em><em>/</em><em>cm³</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em> </em><em>1</em><em>0</em><em>0</em><em>0</em><em> </em><em>Kg</em><em>/</em><em> </em><em>m³</em>

<em>acceleration</em><em> </em><em>due</em><em> </em><em>to</em><em> </em><em>gravity</em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>m</em><em>/</em><em>s²</em>

<em>Height</em><em> </em><em>=</em><em> </em><em>5</em><em>0</em><em> </em><em>cm</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>0</em><em>.</em><em>5</em><em> </em><em>m</em>

after collecting the data, use the formula to solve

<em>pressure</em><em> </em><em>=</em><em> </em><em>pgh</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em>0</em><em>0</em><em> </em><em>×</em><em> </em><em>1</em><em>0</em><em> </em><em>×</em><em> </em><em>0</em><em>.</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>5</em><em>0</em><em>0</em><em>0</em><em> </em><em>Pa</em>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em>

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Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
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The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

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2 years ago
How can you measure the distance an object has moved?
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Can the tangent line to a velocity vs. time graph ever be vertical? Explain.
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hope this helps

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3 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

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=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

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