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tatyana61 [14]
2 years ago
7

Consider a very small hole in the bottom of a tank 17.0 cm in diameter filled with water to a heightof 90.0 cm. Find the speed a

t which the water exits the tank through the hole.20)A) 44.1 m/s B) 48.3 m/s C) 4.20 m/s D) 17.64 m/s
Physics
1 answer:
umka21 [38]2 years ago
5 0

Answer:

Speed of water, v = 4.2 m/s

Explanation:

Given that,

Diameter of the tank, d = 17 cm

It is placed at a height of 90 cm, h = 0.9 m

We need to find the speed at which the water exits the tank through the hole. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 0.9}

v = 4.2 m/s

So, the speed of water at which the water exits the tank through the hole is 4.2 m/s. Hence, this is the required solution.

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I think it might be heat energy. light transforms into heat energy
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2 years ago
You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond
luda_lava [24]

Answer:

a) L = 0.75m   f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m   f₁ = 56.67 Hz ,  f₃ = 170 Hz

Explanation:

This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung

In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1  node         λ₁ = 4 L

With 2 nodes      λ₂ = 4L / 3

With 3 nodes       λ₃ = 4L / 5

The general term would be      λ_n= 4L / n         n = 1, 3, 5, ((2n + 1)

The speed of sound is

         v = λ f

         f = v / λ

         f = v  n / 4L

Let's consider each length independently

L = 0.75 m

        f₁ = 340 1/4 0.75 = 113.33 n

         f₁ = 113.33 Hz

        f₃ = 113.33   3

       f₃ = 340 Hz

L = 1.5 m

       f₁ = 340 n / 4 1.5 = 56.67 n

       f₁ = 56.67 Hz

       f₃ = 56.67 3

       f₃ = 170 Hz

8 0
2 years ago
A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
2 years ago
2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia
Elanso [62]

Answer:

Explanation:

The specific heat of gold is 129 J/kgC

It's melting point is 1336 K

It's Heat of fusion is 63000 J/kg

Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,

The first is E1 = 63000 J/kg x 1.5 = 94500 J

the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

Therefore, all solid is not correct. You will have a mixture of solid and liquid.

For more detail, the difference between E1 and E2 is 7812 J, and that will melt

7812/63000 = 0.124 kg of the solid gold

8 0
2 years ago
(a). The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 631m
lions [1.4K]

Explanation:

Given that,

Length of the building, l = 631 m

Breadth of the building, b = 707 yards

Height of the building, h = 110 ft

1 meter = 3.28084 feet

631 m = 2070.21 feet

1 yard= 3 feet

707 yards = 2121 feet

(a) The volume of any cuboidal shape is given by :

V=l\times b\times h

V=2070.21\ ft\times 2121\ ft\times 110\ ft

V=4.83\times 10^8\ ft^3

(b) V=4.83\times 10^8\ ft^3

V=(4.83\times 10^8\ ft^3)(\dfrac{1\ m}{3.281\ ft})^3

V=1.37\times 10^7\ m^3

Hence, this is the required solution.

3 0
2 years ago
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