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3 years ago

What is the pressure at the bottom of a 2.50 m deep pool of water?

Physics

1 answer:

Grace [21]3 years ago

8 0

Answer:

Pressure = Density x Gravity acceleration x Height

Density of water = 1000 Kgm^-3

Gravity acceleration = 9.8 ms^-2

So it is 4.50.

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What is the relationship between the structure of a white blood cell and its function in the body.

mario62 [17]

White blood cell,also called leukocyte or white corpuscle,a cellular component of the blood that lacks hemoglobin,has a nucleus,is capable of motility,and defends the body against infection and disease by ingesting foreign materials and cellular debris,by destroying infectious agents and cancer cells.

6 0

3 years ago

Which factor indicates the amount of charge on the source charge?

Elina [12.6K]

B. The number of field lines on the source charge.

4 0

3 years ago

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A uniform meterstick in static rotational equilibrium when a mass of 220 g is suspended from the 5.0 cm mark, a mass of 120 g is

KatRina [158]

Answer:

m = 170 g

Explanation:

Meter stick is suspended at 40 cm mark

So here the torque due to additional mass and torque due to weight of the spring must be counter balanced

given that

1) 220 g is suspended at x = 5 cm

2) 120 g is suspended at x = 90 cm

3) mass of the scale is acting at its mid point i.e. x = 50 cm

now with respect to the suspension point the torque must be balanced

so we have

$220(40 - 5) = m(50 - 40) + 120(90 - 40)$

$220(35) = 10 m + 6000$

by solving above equation we have

m = 170 g

7 0

4 years ago

a parallel plate capacitor has square plates that have edge length equal to 100cm and are separated by 1 mm. It is connected to

Flura [38]

Answer:

the energy is stored in the capacitor is 0.32 μJ

Explanation:

Given;

distance of separation, d = 1 mm = 0.001 m

edge length of the square = 100 cm

potential difference across the plates, V = 12 v

let the side of the square = L

This edge length is also the diagonal of the square which makes a right angle with the side of the square.

Applying Pythagoras theorem;

L² + L² = 100²

2L² = 100²

L² = 100²/2

Note area of a square is L²

A = L² = 100²/2 = 5000 cm²

A (m²) = 5000 cm² x 1m²/(100 cm)²

A = 5000 cm² x 1m²/10000 cm²

A = 0.5 m²

Energy stored in a parallel plate capacitor, E= ¹/₂CV²

C = ε₀A/d

where;

ε₀ is permittivity of free space = 8.85 x 10⁻¹² F/m

d is the distance of separation = 0.001 m

A is the area of the plate

C = ε₀A/d = (8.85 x 10⁻¹²)x0.5 / 0.001

C = 4425 x 10⁻¹² F

E = ¹/₂CV² = ¹/₂ x 4425 x 10⁻¹² x ( 12)²

E = 318600 x 10⁻¹² = 0.32 μJ

Therefore, the energy is stored in the capacitor is 0.32 μJ

8 0

3 years ago

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A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

AVprozaik [17]

To solve this problem it is necessary to take into account the concepts related to the magnetic moment and the torque applied over magnetic moments.

For the case of the magnetic moment of a loop we have to,

$\mu = IA$

Where

I = Current

A = Area of the loop

Moreover the torque exerted by the magnetic field is defined as,

$\tau = IAB$

Where,

I = Current

A = Area of the loop

B = Magnetic Field

PART A) First we need to find the perimeter, then

$P = 2\pi r$

$r = \frac{P}{2\pi}$

$r = \frac{1.9}{2\pi}$

$r = 0.3025m,$

The total Area of the loop would be given as,

$A = \pi r^2$

$A = \pi 0.3025^2$

$A = 0.287m^2$

Substituting at the equation of magnetic moment we have

$\mu = (16*10^{-3})(0.287)$

$\mu = 4.58*10^{-3} A.m^2$

Therefore the magnetic moment of the loop is $4.58*10^{-3}Am^2$

PART B) Replacing our values at the equation of torque we have that

$\tau = IAB$

$\tau = (16*10^{-3})(0.287)(0.790)$

$\tau = 3.62*10^{-3}Nm$

Therefore the torque exerted by the magnetic field is $3.62*10^{-3}Nm$

6 0

3 years ago