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1 year ago

True or FalseIf an object is submerged in water, on earth, at a depth of 3km, the Pressure on the object is2.94 x 10^7 Pa

Physics

1 answer:

zhenek [66]1 year ago

6 0

The pressure of a submerged object in a fluid is given by:

$P=\rho gh+P_{atm}$

where ρ is the density of the fluis, g is the acceleration of gravity, h is the depth of the object and Patm is the pressure of the atmosphere. In this case we know that:

• The density of water is 1000 kg/m^3

• The acceleration of gravity is 9.8 m/s^2

• The depth of the object is 3 km, that is, 3000 m.

• The atmospheric pressure is 101325 pascals.

Plugging these values in the equation given above we have:

$\begin{gathered} P=(1000)(9.8)(3000)+101325 \\ P=2.95\times10^7 \end{gathered}$

Therefore, the pressure at this depth is 2.95x10^7 Pa.

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2 years ago

A magnifying glass or a hand lens is a bi-convex lens. It is convex on both sides, meaning that the glass is curved outward to

tankabanditka [31]

Answer:

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3 years ago

Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They

timama [110]

Answer:

c. 48 cm/s/s

Explanation:

Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with a mass of 2M when acted upon by a net force of 2F?

from newtons second law of motion ,

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we can say that

f=m(v-u)/t

a=acceleration

t=time

v=final velocity

u=initial velocity

since a=(v-u)/t

f=m*a

force applied is F

m =mass of the object involved

a is the acceleration of the object involved

f=m*48.........................1

in the second case ;a mass of 2M when acted upon by a net force of 2F

f=ma

a=2F/2M

substituting equation 1

a=2(M*48)/2M

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6 0

3 years ago

The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of

kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

$P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}$

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

$=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m$

Δh_air=714m

7 0

3 years ago

What is the angle of the 2nd order bright fringe produced by two slits that are 8.25x10-5m apart if the wavelength of the incide

dexar [7]

In order to calculate the angle, we can use the formula below for a constructive interference (the interference is constructive because the fringe is bright):

$d\sin\theta=m\lambda$

Where d is the distance between the slits, m is the order of the interference and lambda is the wavelength.

So, using d = 8.25 * 10^-5, m = 2 and lambda = 4.5 * 10^-7, we have:

$\begin{gathered} 8.25\cdot10^{-5}\cdot\sin\theta=2\cdot4.5\cdot10^{-7}\\ \\ \sin\theta=\frac{9\cdot10^{-7}}{8.25\cdot10^{-5}}\\ \\ \sin\theta=1.091\cdot10^{-2}\\ \\ \theta=0.625° \end{gathered}$

Therefore the correct option is the second one.

5 0

1 year ago