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2 years ago

6

A scuba diver measures an increase in pressure of around 10^5 Pa upon descending by 10 m, what is the change in force per square

centimetre on the diver's body?

1 answer:

IceJOKER [234]2 years ago

7 0

Answer:

Explanation:

Given that, .

Pressure around scuba is

P = 10^5 Pa

1 Pa = 1 N/m²

Then

P = 10^5 N/m²

Descending height

h = 10m

Change in force per unit square centimetre

We know that,

Pressure = Force / Area

Then,

Force / Area is the required question we are finding

Then,

Force / Area = 10^5 N / m²

So, let convert the m² to cm²

100cm = 1m

(100cm)² = (1m)²

10⁴cm² = 1m²

Then,

Force / Area = 10^5 N/m² × 1m² / 10⁴cm²

Force / Area = 10 N/cm²

So, the force per unit square centimeters is 10.

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3 years ago

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Answer:

Nature selects organisms with specific heritable traits to survive and reproduce

Explanation:

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For example, let's say that there is a boy that hurt by cats and it's ingrained in his head that cats possess high level of danger to him. Even after that boy grow up, his subsconcious would most likely cause a certain level of paranoia that make him either scared of cats or simply annoyed by seeing them.

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3 years ago

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2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$ , $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

$B = 5.33\times10^{-5} T$

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3 years ago

A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

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2 years ago