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tatyana61 [14]
3 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]3 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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A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f
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Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

W_f = F_fs = 15*0.5 = 7.5 J

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

E_p = mgh = 0.1*10*1.5 = 1.5 J

Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

6 0
3 years ago
11) Which statement about covalent bonds is true? A) Covalent bonds are generally weaker than ionic bonds because they form full
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C) Covalent bonds are generally weaker than ionic bonds because they overlap electrons to fill their outer shell.

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3 years ago
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A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are rele
ZanzabumX [31]

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

u^2-u'^2 = 2as

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

u^2-0^2 = 2\times9.8\tiimes1.08

or

u = \sqrt{21.168}

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed (v_f) = v = 4.6 m/s

Then conservation of energy says  

\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2  

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s  

u₂ = velocity of the basketball before collision= 4.6 m/s  

v₁ =  velocity of the tennis ball after collision  

v₂ = velocity of the basketball  after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2

substituting the values in the above equation we get

v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6

or

v = -8.565-3.965

or

v = -12.53m/s

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

K.E = \frac{1}{2}mv^2

or

K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

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3 years ago
How much positive and negative charge is there in cup of water?
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In one cup of water (250 ml) there are 1.338×10 to the 7th power C positive charge and - 1.338×10 to the 7th power C negative charge. Totally they will be neutral
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