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joja [24]
2 years ago
15

What level of education is required to obtain a position as a insurance billing specialist ?

Physics
2 answers:
Georgia [21]2 years ago
8 0

Answer:

Billing experts at the entry level are normally expected to have a high school diploma or equivalent, as well as skills, certifications, and experience. Some jobs require at least an associate's or bachelor's degree.

Explanation:

nika2105 [10]2 years ago
4 0

Answer: high school diploma

Explanation:

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Which celestial body would have the strongest gravitational pull on a satellite orbiting 100 km above its surface? Explain your
Andreyy89

Answer:

Jupiter

Explanation:

Since the mass of Jupiter is the greatest from the given choices, it will exert the most force on any object orbiting 100km above its surface.

This is compliance with the Newton's law of universal gravitation which states that "the force of attraction between two bodies is directly proportional to the magnitude of their masses and inversely proportional to the distances between them".

  • Therefore, the more the masses of two bodies, the higher the gravitational attraction
  • Since the distance is the same, the planet with the greater mass will exert the most force on the satellite.
3 0
3 years ago
Read 2 more answers
In a pulley system, two blocks are connected by a rope as shown below. The coefficient of static friction between block A and th
umka21 [38]

Hi there!

We can begin by doing a summation of forces for each block.

Block A:

This block has the force of tension (in direction of acceleration +) and static friction (opposite direction -) acting on it. Thus:

\Sigma F_A = T - F_s\\\\m_Aa = T - \mu m_Ag

Block B:

This block has the force of tension (opposite of acc. -) and gravity (in direction of acc +), working on it.

\Sigma F_B = m_Bg - T\\\\ m_Ba = m_Bg - T

Add both of the expressions and solve for the maximum mass of Block B.

\Sigma F = m_Bg - T + T - \mu m_Ag\\\\a(m_A + m_B) = m_Bg - \mu m_Ag

To find the minimum value, we can set a = 0, so:

0 = m_Bg - \mu m_Ag\\\\\mu m_Ag = m_B g\\\\0.8(20)(9.8) = m_B (9.8)\\\\m_B = \frac{0.8(20)(9.8)}{9.8)} = \boxed{16 kg}

The block must weigh <u>> 16 kg</u> for block A to move.

5 0
3 years ago
An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop the loop maneuver. The acceleration
Natali5045456 [20]

The radius of the loop is 18.9 km

Explanation:

When the airplane is at the top of the loop, the pilot experiences two forces:

  • The force of gravity, acting downward, of magnitude mg
  • The normal reaction exerted by the seat on the pilot, also acting  downward, N

Since the plane is moving in a circular motion, the net force on the pilot must be equal to the centripetal force, therefore we can write:

mg+N = m\frac{v^2}{r}

where

m is the mass of the pilot

g=9.8 m/s^2 is the acceleration of gravity

N is the normal reaction

v = 430 m/s is the speed of the plane

r is the radius of the loop

Here we are told that the pilot feel weightless at the top of the loop: this means that the normal reaction is zero,

N = 0

Therefore the equation becomes

mg=\frac{mv^2}{r}

And so we can find the radius of the loop:

r=\frac{v^2}{g}=\frac{430^2}{9.8}=18.9 \cdot 10^3 m = 18.9 km

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0
3 years ago
Electrons in an x-ray tube are accelerated through 144 kV and directed toward a target to produce x-rays. Calculate the power of
Katarina [22]

Answer:

Power of the electron beam in the x-ray tube is 2304 W

Explanation:

Given;

voltage of the electron beam in the x-ray tube, V =  144 kV = 144 x 10³ V

current of the electron beam in the x-ray tube, I = 16.0 mA = 16 x 10⁻³ A

Power is given as the product of voltage and current in a circuit.

Power of the electron beam in the x-ray tube, P = V x I

Power of the electron beam in the x-ray tube, P = 144 x 10³ V x 16 x 10⁻³ A

P = 2304 W

Therefore, power of the electron beam in the x-ray tube is 2304 W

7 0
3 years ago
How much would a 70.0-kg person weigh on Mercury?
Agata [3.3K]

AS Weight = w =mg

          m= 70kg

 g on mercury =3.7 m/s²

so  W = (70kg) ( 3.7m/s²)

      w=259 kgm/s²

w = 259 N

3 0
4 years ago
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