2. Why do you think a light stick would be a good light source around flammable material?

Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun. The answer is option C

According to Kepler's third law of planetary motion for planets and satellites, he described circular orbit about the same center body stating that:

The squared of the period is directly proportional to the cube of the radius. That is,

[tex]T^{2} \alpha r^{3}[/tex]

The orbital period formula is expressed as :

[tex]T^{2}[/tex] = 4[tex]\pi ^{2}[/tex] [tex]r^{3}[/tex]  ÷ GM

Suppose a planet orbits a star at the same distance as Earth (1 AU) but with a period of roughly 200 Earth days. The mass of the central star will be greater than the mass of our sun based on the formula given above.

In a geostationary orbit, whatever the mass of the satellite may be, they will travel at the same speed in a particular orbit. mass M in the above formula becomes the mass of the sun or the center mass.

Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun.

Therefore, the correct answer is option C

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Answer:

Two positively charged particles

Explanation:

I said two positively charged particles because if I say c or d what ever it is for you guy it can be wrong so just pick the one that says Two positively charged particles

Answer: C  Two positively charged particles.

Explanation:

just trust me on it, i've done the same quiz.

Answer: add it

Explanation:

Answer:

The sum of vectors P and Q is given by the vector R, the resultant sum vector using the parallelogram law of vector addition. If the resultant vector R makes an angle ϕ with the vector P, then the formulas for its magnitude and direction are: |R| = √(P2 + Q2 + 2PQ cos θ) β = tan-1[(Q sin θ)/(P + Q cos θ)]

Ectoparasites are a taxonomically diverse group of organisms that infest the skin of human beings and other animals. Ectoparasitic arthropods and nematodes are similar in that an individual organism can produce skin lesions that are large enough to see with the unaided eye. Ectoparasitic infestations are often intensely itchy, causing considerable annoyance and discomfort. These conditions are often focally hyperendemic in impoverished communities, with a particularly high prevalence in vulnerable families, households, and neighborhoods.

Pediculosis (infestation by head and body lice) and scabies are found to some degree in all human populations, but myiasis (fly larva infestation), tungiasis (sand flea disease), and cutaneous larva migrans occur mainly in tropical and subtropical environments. Except for body lice, the organisms discussed in this article are not vectors of pathogenic microorganisms. In other words, most ectoparasites do not carry disease-causing agents; they are, instead, the direct cause of disease. Mortality is low, but the cumulative morbidity from the direct discomfort, secondary bacterial infections, and sequelae of those infestations and infections is considerable.

Despite the abundant presence of ectoparasitic infestations across human populations, biomedical science lacks firm evidence-based practices to reliably control these organisms. In addition, head lice and scabies are developing resistance to some chemical compounds employed to treat infested individuals, prevent spread, and control outbreaks.

Please mark me brainliest.

The magnitude of the magnetic field B(r) inside the capacitor as a function of distance from the axis is [tex]\mathbf{B(r) =\dfrac{\mu_o I r}{2 \pi a^2} }[/tex]

A charged capacitor usually cause changes between the plates of electric field E and the electric flux Φ

As seen in Ampere's law which is extended by Maxwell:

[tex]\mathbf{\oint B^{\to} .dl^{\to} = \mu_o \Big( I+\dfrac{d \phi }{dt} \Big)}[/tex]

To meet the equilibrium equation of electric charge, Maxwell modified Ampere's law by incorporating the displacement current through into electric current component.

∴

The displacement current can be expressed by the relation;

[tex]\mathbf{I = \varepsilon_o \dfrac{dE}{dt} }[/tex]

From the given information, the displacement current density through area A can now be expressed as:

[tex]\mathbf{I = \varepsilon_o A \dfrac{dE}{dt} }[/tex]

Replacing the value of displacement current density into Maxwell modified Ampere Law, we have:

[tex]\mathbf{\oint B .dl= \mu_o I_{enclosed} }[/tex]

[tex]\mathbf{B \times (2 \pi r) = \mu_o I\Big (\dfrac{\pi r^2}{\pi a^2} \Big) }[/tex]

[tex]\mathbf{B(r) =\dfrac{\mu_o I r}{2 \pi a^2} }[/tex]

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Answer:

120km

Explanation:

To get speed you divide distance by time.

Answer:

Alice works with natural rubber and Julio works with vulcanised rubber.

Explanation:

Answer:

B. Alice works with natural rubber, and Julio works with vulcanized rubber.

Explanation:

I just took the test and got 100 percent.

I hope this helps you!!!!

Please mark my answer the Brainiset. Thank You!!!!!

Explanation:

All motions are relative to some frame of reference. Saying that a body is at rest, which means that it is not in motion, merely means that it is being described with respect to a frame of reference that is moving together with the body

Answer:

Motion is always described in relation to a reference point.

Explanation:

For example, if a car is driving at the speed of 30 miles per hour and you are standing by the road, to you, the car appears to be moving 30 miles per hour down the road. However, if you were in a car also driving at 30 miles per hour next to the truck, the same truck wouldn't appear to move.

Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.

Explanation: Hope this helped :)

Answer:

• Average current = 25,000 Amperes

• Time = 0.80 seconds

a)

[tex]{ \rm{quantity \: of \: charge = current \times time}} \\ { \rm{q = 25000 \times 0.80}} \\ { \boxed{ \mathfrak{answer : \: { \rm{quantity \: of \: charge = 20000 \: coulombs}} }}}[/tex]

b)

[tex]{ \rm{number \: of \: electrons = \frac{current}{charge \: quantity} }} \\ \\ { \rm{n = \frac{25000}{20000} }} \\ \\ { \boxed{ \mathfrak{ \: answer : \: { \rm{n = 1.25 \: }}}}}[/tex]

[tex]{ \underline{ \mathfrak{ \green{Good \: grace}} \: \:⚜ \: \: { \blue{ \mathfrak{Good \: God}}}}}[/tex]

Answer:

60 meters

Explanation:

If you are going 3 meters in a second, and you are traveling for 20 seconds, you have to multiply

3meters/second*20seconds

cross out the seconds and you have

3 meters*20

60 meters

Answer:

I think it's true,,

Explanation:

maybeee truee

Answer:

the moving of something from its place or position

or

he occupation by a submerged body or part of a body of a volume which would otherwise be occupied by a fluid

There are 2 different definitions.

1. The moving of something from its place or position.

2. The occupation by a submerged body or part of a body of a volume which would otherwise be occupied by a fluid.

Answer:  his total speed would be 3 seconds

Explanation: because if you mulimply 10 x 3 you get 30 meaning that your answer would be 3

Answer:

A - Juan

Explanation:

I took the test!

Please find attached photographs for your answer.

Hope it helps.

Do comment if you have any query.

Answer:

2400m/s , 2240,2080,1920m/s

Answer:

Sorry I need points [tex] \boxed{}[/tex]

Answer:

[tex]V_{2.2} =2.2V[/tex]

Fast and loose: that's a classic voltage divider. the drop from the i-th resistor is [tex]R_i \over {\sum R}[/tex] of the drop across the whole series. In our situation, it's [tex]2.2\cdot 10^3 \over {2.2\cdot10^3 + 6.8\cdot 10^3[/tex] of 9 V. By plugging numbers in a calculator, it's 22/90 of 9V, or 2.2V

With Ohm's Law

The series is equivalent of a single resistor of Resistance [tex]2.2\cdot 10^3+6.8\cdot 10^3 \Omega = 9.0 k\Omega[/tex]. By Ohm's first Law ([tex]V=Ri[/tex]) the current flowing through the resistor is [tex]9V = 9*10^3\Omega i \rightarrow i=1mA[/tex]. At this point, the drop across the first resistor is, again by Ohm's law[tex]V = 2.2 \cdot 10^3 \Omega \cdot 1\cdot 10^{-3} A = 2.2V[/tex]

Answer:

"H" is a chemical symbol and "H2" is chemical formula.

Answer:

It is important to use renewal resources because we use so many non- renewable resources that we may lose the ability to make a lot of things like gasoline or power for our cities.

Answer:

The rocket travelled 37km per minute

Explanation:

185÷5=37

Answer:

The entire electromagnetic spectrum, from the lowest to the highest frequency (longest to shortest wavelength), includes all radio waves (e.g., commercial radio and television, microwaves, radar), infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

The given kinematic equations of motion are derived from Newton's Laws

of Motion.

Reasons:

The direction the ball is kicked = Upwards in the air

Maximum height to which the ball rises = 6.25 m

A) The speed of the ball the instant it left the soccer player's foot

Solution:

The equation to use is; v² = v₀² + 2·[tex]a_y[/tex]·(y - y₀)

[tex]a_y[/tex] = The acceleration due to gravity ≈ -9.81 m/s²

At the maximum height, y - y₀ = 6.25 m

The velocity at the maximum height, v = 0

v₀ = The speed of the ball the instant it left the soccer player's foot

Which gives;

v² = v₀² + 2·[tex]a_y[/tex]·(y - y₀)

0 = v₀² - 2 × 9.81 × (6.25) = v₀² - 122.625

v₀² = 122.625

v₀ = √(122.625) ≈ 11.074

The ball's speed the instant it left the soccer player's foot, v₀ ≈ 11.074 m/s.

B) Required:

The time it takes the soccer ball to rise to the maximum height.

Solution:

The equation of motion required is v = v₀ + [tex]a_y[/tex]·t

At the maximum height, v = 0

v₀ ≈ 11.074 m/s

v = v₀ + [tex]a_y[/tex]·t

Which gives;

0 = 11.074 - 9.81 × t

[tex]t \approx \dfrac{11.074}{9.81} \approx 1.129[/tex]

The time it took the ball to rise to its maximum height, t ≈ 1.129 seconds

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Answer:

Fc= 47.363 N

Explanation:

First, you need tangential velocity because they don't give that to you in the problem.

So, the formula for tangential velocity is υ=2πr/t.

For this problem its v=2π4/10= 2.513 m/s

Then we plug this into our equation for centripetal force

Fc= 30*(2.513)^2/4=47.363 N

Fc=47.363 N

The centripetal force acting on Tommy is 47.36 N if Tommy, who has a mass of 30 kg, sits 4.0 meters from the center of a merry-go-round that is rotating for a period of 10 seconds.

A tangible body's mass is the amount of matter it possesses. It's also a metric of inertia or the resistance to velocity when a net force is exerted.

It is given that:

Tommy, who has a mass of 30 kg, sits 4.0 meters from the center of a merry-go-round that is rotating for a period of 10 seconds.

As we know the formula for the centripetal force:

Fc = mv²/r

m = 30 kg

r = 4 meters

First, find the tangential velocity:

The formula for tangential velocity is:

v = 2πr/t

v = 2π4/10

v = 2.513 m/s

Plug the values in the formula:

Fc= 30(2.513)²/4

Fc = 47.36 N

Thus, the centripetal force acting on Tommy is 47.36 N if Tommy, who has a mass of 30 kg, sits 4.0 meters from the center of a merry-go-round that is rotating for a period of 10 seconds.

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