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Neporo4naja [7]
2 years ago
3

Circle Lab

Physics
1 answer:
Delvig [45]2 years ago
3 0

The circumference of a circle is directly proportional to the diameter of the circle.

<h3>What is the relationship between the circumference of the circle and the diameter of the circle?</h3>

The circumference of the circle and the diameter of the circle are directly related by the equation given below:

  • Circumference of a circle = π * diameter

where π is a constant.

Therefore, in the circle lab:

  • The dependent variable is the circumference of the circle
  • The independent variable is the diameter of the circle
  • The shape of the line is a straight line
  • The equation o the line is that of the general equation of a straight line; y = mx + c

<h3>Why did I do this lab?</h3>

The purpose of the lab was to determine the relationship between the circumference of the circle and the diameter of the circle

<h3>What did I learn from doing this lab?</h3>

I have learnt that the circumference of a circle is directly proportional to the diameter of the circle.

<h3>What are some errors that may have occurred when I performed this lab?</h3>

Some errors that occurred include:

  • error in taking readings from the metric ruler
  • error when taking the circumference of the circle with the string.

In conclusion, the circle lab was done to determine the relationship between the circumference and the diameter of a circle.

Learn more about circumference and diameter of a circle at: brainly.com/question/1582885

#SPJ1

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Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and the
MissTica

Answer:

V = 10581.59 V

Explanation:

To solve this problem, we have to take into the relation between the kinetic energy of an electron and the energy generated by the potential difference of an electric field. This relation is given by

E_{k} = E_{V}

\frac{1}{2}mv^{2} = eV

where m is the mass of the electron, v is the velocity of the electron, e is the electric charge and V is the potential difference that accelerated the electron.

By taking V from the last expression we have:

V = \frac{mv^{2}}{2e} = \frac{(9.1*10^{-31} )(6.1*10^{7})^{2}}{2*(-1.6*10^{-19})}} = 10581.59 V

where we have taken that

me = 9.1*10^(-31) Kg

e = -1.6*10^(-19) C

v = 6.1*10^(7)

I hope this is useful for you

Regards

6 0
4 years ago
CAN ANYONE PLEASE HELP ME WITH THIS ONE ITS URGENTLY NEEDED
denis-greek [22]

Answer:

Explanation:

conservation of momentum

initial momentum is zero

Lets say that m₁ moves along the x axis

and m₂ moves along the y axis

in the y direction

0 = m₁(0) + m₂(30) + m₃(vy)

0 = m₁(0) + m₂(30) + 3m₂(vy)

-3m₂(vy) = m₂(30)

vy = -10 m/s

in the x direction

0 = m₁(30) + m₂(0) + m₃(vx)

0 = 2m₂(30) + m₂(0) + 3m₂(vx)

-3m₂(vx) = 2m₂(30)

vx = -20 m/s

v = √(-10² + -20²) = 10√5 m/s ≈ 22.36 m/s

θ = arctan(-10/-20) = 206.56505... ≈ 206.6°  CCW from the + x axis

4 0
3 years ago
List and describe the forces that act on a block as it moves down the ramp at constant speed.
Vitek1552 [10]

gravity, inertia, friction, kinetic energy?


7 0
3 years ago
A 75 kg bungee jumper jumps off with zero initial speed from a high cliff with a bungee cord tied to her ankle. At some time dur
stich3 [128]

Answer:

The elastic potential energy PE=143.47kJ

Explanation:

This problem bothers on the potential energy stored in a material.

Given data

Mass of the bungee jumper

m= 75kg

Height of jump down the cliff h=195m

We know that the elastic potential energy stored is same as the potential energy of the bungee jumper

PE= 1/2kx²= mgh

Assuming g is 9.81m/s²

PE= 75*9.81*195

PE= 143471.25J

PE=143.47kJ

What is potential energy?

Potential energy is the energy possessed by a body by virtue of it position.

4 0
3 years ago
How does the orbital speed of an asteroid in a circular solar orbit with a radius of 4.0 AU compare to a circular solar orbit wi
Murrr4er [49]

Answer:

The circular solar orbital speed at 4.0AU is 1/4( one fourth) that at 1.0AU

Explanation:

am = mvr= angular momentum

am4= 4mvt

am1= mvp1

Vt=1/4vp

Vp=4vt

am1= 4mvt

am1=am4

The circular solar orbital speed at 4.0AU is 1/4 (one fourth) that at 1.0AU

6 0
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