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2 years ago

7

Circle O has a circumference of approximately 44 in. Circle O with diameter d is shown. What is the approximate length of the di

ameter, d?

2 answers:

Zanzabum2 years ago

7 0

The answer is 14.001

GrogVix [38]2 years ago

5 0

Answer:

14.001 in

Explanation:

Circumference = pi * diameter

Circumference / pi = diameter

44 inches/pi = 14.001 inches

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A ball is swung in a horizontal circle at a constant speed. Each circle takes 0.85 seconds to complete and the rope is 0.40 m lo

BaLLatris [955]

Answer:

The centripetal acceleration will be "21.785 m/s²".

Explanation:

The given values are:

Time,

t = 0.85 seconds

Length of rope,

r = 0.40 m

Mass of ball,

m = 0.80 kg

As we know,

⇒ $w=\frac{2 \pi}{t}$

On substituting the values, we get

⇒ $=\frac{2\times 3.14}{0.85}$

⇒ $= \frac{6.28}{0.85}$

⇒ $=7.38 \ rad/s^2$

The centripetal acceleration will be:

⇒ $a=r\times w^2$

⇒ $=0.40\times (7.38)^2$

⇒ $=0.40\times 54.46$

⇒ $=21.785 \ m/s^2$

7 0

2 years ago

A person's prescription for bifocals is -0.25 diopter for distant vision and +2.50 diopters for near vision, the near vision bei

Darina [25.2K]

Answer:

net power is + 2.25 D

Explanation:

Given data

distance vision = -0.25 D

near vision = + 2.50 D

to find out

net power

solution

we have given a person lens power for near is - 0.25 diopter and lens power for near power is +2.50 diopter so

net power is sum of both the power vision

so

net power = distance + near power

put both value we get net power

net power = ( -0.25 D) + ( + 2.50 D)

net power = + 2.25 D

so net power is + 2.25 D

6 0

3 years ago

50 points!!! Kinetics

iris [78.8K]

Answer:

98 m √

Explanation:

How about s = Vo * t + ½at² ?

s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6

and

h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490

Subtract 2nd from first:

0 = -8Vo + 470.4

Vo = 58.8 m/s

h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m

3 0

2 years ago

Read 2 more answers

Hey can someone send me the answer to this

bagirrra123 [75]

Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.

5 0

3 years ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

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F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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3 years ago