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3 years ago

How to measure the external diameter of a sphere

1 answer:

4 0

Sphere is that the circular objects in the two dimensional space (1) circle
(2) disk. Two dimensional space is a set of points and the distance of that point,The two points of Sphere that length and center.
Sphere can constructed as the named of surface form circle about any diameter. circle is the special type of the revolution replacing the circle,
sphere is the distance r is the radius of the ball and circle is the center of mathematical ball,as the center and the radius of the sphere is to respectively.
The ball and sphere has not be maintained mathematical references as a solid references. A sphere of any radius is centered at the number of zero.

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The town hall building is situated close to Boojho’s house. There is a clock on the top of the town hall building which rings th

KiRa [710]

Answer:

At night we experience quietness more. At this time, sounds of object will be heard more clearer

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3 years ago

Without fluid friction, all objects accelerate at?

lys-0071 [83]

Answer:

Acceleration of gravity= $9.8m/s/s$

Explanation:

Newton's Second Law-acceleration is proportional to the net force acting on an object.

All objects usually free fall at the same acceleration of $9.8m/s/s$ -this regardless of their mass. This acceleration is known as acceleration of gravity.

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3 years ago

A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga

tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₂² v₂

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{2.25^2}{3.75^2} v_2$

$v_1= 0.36 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)$

$32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)$

$v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}$

$v_2=\sqrt{16.084}$

v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)² x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.