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valentina_108 [34]
3 years ago
15

Imagine you are holding a small ball below you so that you look down on its "north pole." turn the ball in a counterclockwise di

rection. continue to turn it in that direction as you slowly raise it above your head so that you are looking up at its "south pole." in what direction is it rotating now?
Physics
1 answer:
enyata [817]3 years ago
8 0
It is rotating Clockwise
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Atoms with the same atomic number but different atomic mass are called
Alexxx [7]
<span>Atoms with the same atomic number but different atomic mass are called:

<span>Isotopes</span>
</span>
4 0
2 years ago
Read 2 more answers
A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. What is the
iren2701 [21]

Answer:

a)  T = 1,467 s , b)    A = 0.495 m , c)  v = 4.97 10⁻² m / s

Explanation:

The simple harmonic movement is described by the expression

        x = A cos (wt + Ф)

Where the angular velocity is

       w = √ k / m

a) Ask the period

Angular velocity, frequency and period are related

      w = 2π f = 2π / T

      T = 2π / w

      T = 2pi √ m / k

      T = 2π √ (1.2 / 22)

      T = 1,467 s

      f = 1 / T

      f = 0.68 Hz

b) ask the amplitude

The mechanical energy of a harmonic oscillator

        E = ½ k A²

       A = √2 E / k

       A = √ (2 2.7 / 22)

       A = 0.495 m

c) the mass changes to 8.0 kg

As released from rest Ф = 0, the equation remains

         x = A cos wt

        w = √ (22/8)

        w = 1,658

         x = 3.0 cos (1,658 t)

Speed ​​is

         v = dx / dt

         v = -A w sin wt

The speed is maximum when without wt = ±1

         v = Aw

         v = 0.03    1,658

         v = 4.97 10⁻² m / s

6 0
2 years ago
a steel ball bearing is released from a height hhh and rebounds after hitting a steel plate to a height hhh. what is true about
lbvjy [14]

In collision of the steel ball and the steel plate, the collision is an inelastic collision and there is loss in the kinetic energy.

<h3>What are collisions?</h3>

Collisions occur when two objects that are moving in the same directions or in different direction meet each other and collide.

There are two types of collisions:

  • elastic collision - the kinetic energy is conserved
  • inelastic collision - there is a loss in kinetic energy

In the collision of the steel ball and the steel plate, there is loss in the kinetic energy of the steel ball which is converted to sound energy.

In conclusion, the collision of the steel and steel plate is an inelastic collision.

Learn more about collisions at: brainly.com/question/7694106

#SPJ1

4 0
1 year ago
If you shine a light of frequency 375hz on a double slit setup , and you measure the slit separation to be 950 nm and the screen
amm1812

Answer:

y = 33.93 10⁵ m

Explanation:

This is an interference exercise, for the contributory interference is described by the expression

           d sin θ = m λ

let's use trigonometry for the angle

           tan θ = y / L

how the angles are small

          tan θ = sin θ / cos tea = sin θ

 

we substitute

         sin θ = y / L

        d y / L = m λ

         y = m λ L / d

the light fulfills the relation of the waves

       c = λ f

       λ = c / f

       λ = 3 10⁸ /375

       λ = 8 10⁵ m

first order m = 1

let's calculate

        y = 1  8 10⁵  4030 10-9 / 950 10-9

        y = 33.93 10⁵ m

6 0
2 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
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