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Inessa [10]
4 years ago
5

A bug walks exactly halfway around the edge of a circular cupcake with a diameter of 5 cm what is the distance he traveled and w

hy
Physics
1 answer:
adelina 88 [10]4 years ago
7 0

The circumference of a circle is (pi) x (diameter)

The circumference of the cupcake is (pi) x (5 cm)

Halfway around is (1/2) x (pi) x (5 cm) = (2.5 pi cm) = <em>about 7.85 cm</em>

The 'why' appears up above, in the first 2 lines of this solution.

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Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birm
forsale [732]

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

4 0
3 years ago
at location a, what are the directions of the electric fields contributed by the electron. calculate the magnitudes of the elect
Lisa [10]

We can use the equation E = k | Q | r 2 E = k | Q | r2 to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge,

<h3>What is electric and magnetic field ?</h3>

With the use of electricity and other types of artificial and natural illumination, invisible energy fields known as electric and magnetic fields (EMFs) and radiation are created.

  • While the magnetic field is discernible by the force it exerts on other magnetic particles and moving electric charges, the electric field is actually the force per unit charge experienced by a non-moving point charge at any given location inside the field.

Learn more about Electromagnetic field here:

brainly.com/question/14372859

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4 0
1 year ago
A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is a
Licemer1 [7]

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

8 0
3 years ago
The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an
Readme [11.4K]
1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
substituting the coordinates of the two charges, we get
d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m

2) Then, we can calculate the electrostatic force between the two charges q_1 and q_2, which is given by
F=k_e  \frac{q_1 q_2}{d^2}
where k_e=8.99\cdot10^{9} Nm^2C^{-2} is the Coulomb's constant.
Substituting numbers, we get 
F=8.99\cdot10^{9} Nm^2C^{-2}  \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.
7 0
3 years ago
A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. W
lianna [129]

Answer:

So coefficient of kinetic friction will be equal to 0.4081

Explanation:

We have given mass of the block m = 0.5 kg

The spring is compressed by length x = 0.2 m

Spring constant of the sprig k = 100 N/m

Blocks moves a horizontal distance of s = 1 m

Work done in stretching the spring is equal to W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J

This energy will be equal to kinetic energy of the block

And this kinetic energy must be equal to work done by the frictional force

So \mu mg\times s=2

\mu\times  0.5\times 9.8\times 1=2

\mu =0.4081

So coefficient of kinetic friction will be equal to 0.4081

5 0
3 years ago
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