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3 years ago

Explain the method to measure the external diameter of a sphere

Physics

1 answer:

Serjik [45]3 years ago

8 0

Answer:

Sphere is a geometrical object in dimensional space that surface of a circle and ball.

Explanation:

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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A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the wate

Vilka [71]

Answer:

f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

∑ F = ma

where the acceleration is

a = $\frac{d^2 y}{dt^2 }$

B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

B = W

In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains

B’= m $\frac{d^2 y'}{dt^2 }$

the thrust is given by the Archimedes relation

B = ρ_liquid g V_liquid

the volume is

V = π r² y'

we substitute

- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

$\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0$

this differential equation has a solution of type

y = A cos (wt + Ф)

where

w² = ρ_liquid g π r² /m

angular velocity and frequency are related

w = 2π f

we substitute

4π² f² = ρ_liquid g π r² / m

f = $\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }$

calculate

f = $\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }$

f = 5.3 Hz

6 0

3 years ago

An ice skater is moving in a circle at a constant speed. Which of the following best explains the forces acting on the ice skate

dangina [55]

B because the skater is constantly moving in the same speed to go in a circle .

7 0

3 years ago

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?

jasenka [17]

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

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2 years ago

How did new technology such as the telescope and new theories such as Pascal's Law laid the foundation of the Scientific Revolut

AfilCa [17]

Answer:

"Scientists used them to create new theories"

Explanation:

The Scientific Revolution was a sequence of actions that manifest the development of contemporary science through the early contemporary period, when advances in mathematics, physics, astronomy, biology and chemistry altered the opinions of civilization around nature. The scientific revolution denotes to the quick developments in European scientific, mathematical, and political assumed, grounded on a new philosophy of experimentation and a belief in growth that defined Europe in the 16th and 17th centuries.

4 0

3 years ago

Explain the method to measure the external diameter of a sphere​

Explain the method to measure the external diameter of a sphere