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3 years ago

Hãy kể tên, kí hiệu, đơn vị một số đại lượng Vật Lý mà em đã học ở lớp 10, 11?

Physics

1 answer:

wariber [46]3 years ago

7 0

I’m not Vietnamese but I think your asking for this, for people who understand Vietnamese pls help them :d

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What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p

irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

<em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
<em>Distance between each floor, d = 10 ft</em>

The vertical pressure of the water is calculated as follows;

$P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\$

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10 ft

$P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI$

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

3 0

2 years ago

Two cars go through 2 different crashes. Car 1 experiences a 500 Ns impulse for a duration of 15s, while car 2 experiences that

tatuchka [14]

B is your answer hope this helps;)

5 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago

HELP PLEASE!!

aliya0001 [1]

Answer:

m=41 kg

v=0,02 ms

R=2,1 m

F-?

a=v²/R

a=(0,02) ²/2,1=0,0002

F=m*a

F=41*0,0002=0,0082 H

F=0,0082 H

3 0

3 years ago

Which planet has a tilted axis of rotation similar to that of earth, which means it has seasons?.

MrRa [10]

Answer:

Uranus

Explanation:

it's uranus....

3 0

3 years ago