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Aleks [24]
3 years ago
6

Write this number in standard notation. 4.702 x 10–4

Physics
1 answer:
sergeinik [125]3 years ago
5 0

Answer:

0.0004702

Explanation:

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A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
2 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

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The answer should be flammability
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There are creases. Or just origami
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Taden has found a table of the atmospheric layers of the Sun. However, he can’t make out the names of the layers in the chart.
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The sun is a ball of hot gases containing different kinds of elements at different cores. It has a very high temperature that radiates all throughout the Milky Way galaxy. The sun has three main parts; photosphere, chromospheres and corona. The outer core of a star located at the chromospheres contains mostly of hydrogen. Inside the hydrogen is helium then carbon, oxygen, neon, magnesium silicon and the inert gas. The photosphere is scattered by the loose electrons in the corona’s plasma.

4 0
3 years ago
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