All categories

3 years ago

The circumference of a circle is 12π meters. find the circle diameter.

1 answer:

7 0

The circumference of a circle is 2*pi*r=C so plugging in numbers we have
12pi= 2*pi*r so solving for r gives 12pi/2pi=r which gives 6=r so the radius of the circle is 6 meters since diameter is just 2*r this gives 2*r =12 meters so
Diameter = 12meters

You might be interested in

If an object is projected upward with an initial velocity of 127 ft per? sec, its height h after t seconds is h equals negative

Stolb23 [73]

To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:

h = -16t^2 + 127t

We substitute 55 seconds to t and obtain,

h = -16(55)^2 + 127(55)
h = - 41415

4 0

3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

A scientist carries out an experiment. How could she help other scientists judge the validity of her results?

NNADVOKAT [17]

One way to do it is she could right down the data that she got

4 0

3 years ago

Read 2 more answers

PLEASE HELP THIS IS DUE TODAY!! (GIVING BRAINLIEST TO WHOEVERY ANSWERS QUICKLY AND CORRECTLY!) 15 POINTS!

Jlenok [28]

As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf. ... Like low-mass stars, high-mass stars are born in nebulae and evolve and live in the Main Sequence

hydrogen shell burning - outer layers swell. Red Giant Branch - helium ash core compresses - increased hydrogen shell burning. First Dredge Up - expanding atmosphere cools star - stirs carbon, nitrogen and oxygen upward - star heats up.

5 0

3 years ago

In operant conditioning, generalization occurs when

Vitek1552 [10]

Typically occurs when we associate things to other things that look alike. We see that in many experiments, specifically “Little Albert” who was conditioned to be afraid of rats but later was afraid of anything that resembled that of a rat.
Hope this helps!

5 0

3 years ago

The circumference of a circle is 12π meters. find the​ circle diameter.

The circumference of a circle is 12π meters. find the circle diameter.