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user100 [1]
3 years ago
13

A water tank is filled with water. What will be the pressure of the water when the level of the water is 6m?

Physics
2 answers:
Snezhnost [94]3 years ago
8 0

Answer:

The pressure of the water is 58,800 Kg/s/m² or 58,800 Pa or 58.8 KPa

Explanation:

Height, h = 6 m

Gravity, g = 9.8 ms⁻¹

Density of Water, ρ = 1000 Kgm⁻³

P (fluid) = ρ*g*h

P (fluid) = 1000 x 9.8 x 6 (Kgm⁻³ x ms⁻¹ x m)

P (fluid) = 1000 x 9.8 x 6 (Kgm⁻³ x ms⁻¹ x m)

P (fluid) = 58,800 Kg/s/m²

P (fluid) = 58,800 Pa or 58.8 KPa

Luden [163]3 years ago
3 0

Answer:

pressure = density x g x height

= 1000 x 10 x 6 Pascal

=60000 Pascal

OR 60 kP

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Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0
3 years ago
PLEASE HELP!!!! :D When you look at the Sun through a filtered telescope, the visible portion of the Sun appears blotchy.
olya-2409 [2.1K]
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5 0
3 years ago
Read 2 more answers
true or false/ force field that exists in the space around every massive body is called electric field​
musickatia [10]

Answer:

True

Explanation:

An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

3 0
3 years ago
An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _
Viefleur [7K]
Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
where 
f=5 cm is the focal length
p=10 cm is the distance of the object from the mirror
q is the distance of the image from the mirror.

Rearranging the equation, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}
so, the distance of the image from the mirror is q=10 cm.

1b) The image height is given by the magnification equation:
\frac{h_i}{h_o}=- \frac{p}{q}
where h_i is the heigth of the image and h_o=1 cm is the height of the object. By rearranging the equation and using p and q, we find
h_i=-h_o  \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

3) As before, we find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

And now we can use the magnification equation to find the image height:
\frac{h_i}{h_o}=- \frac{p}{q}
Rearranging it, we find
h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm
and the negative sign means the image is inverted.
5 0
3 years ago
A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w
tatyana61 [14]
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s
And so, the kinetic energy of the object is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
And so the new kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
7 0
2 years ago
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