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antiseptic1488 [7]
3 years ago
7

PLEASE HELP!!! 12-14!!!!!

Physics
1 answer:
kondor19780726 [428]3 years ago
8 0
12-14 equals -2.....
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How to determine the amount of resistance when two or more resistors are in parallel.
Elena L [17]

Answer: 1/R = 1/R1 + 1/R2+ ...+ 1/Rn

R is resistance of system in which there are resistors R1, R2 , ... Rn parallel.

3 0
3 years ago
Which statements describe the image produced by a concave lens?check all that apply.
kotegsom [21]

<em>Answer:</em>

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  • <em>A-which is the image is always right side up.</em>
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7 0
3 years ago
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An arrow of mass 20 g is shot horizontally into a bale of hay, striking the hay with a velocity of 60 m/s. It penetrates a depth
goblinko [34]

From the question, The kinetic energy of the fired arrow is equal to the work done by the bale of hale in stopping the arrow.

We make use of the following formula

mv²/2 = F'd................... Equation 1

Where

  • m = mass of the arrow
  • v = velocity of the arrow
  • F' = average stopping force acting on the arrow
  • d = distance of penetration

Make F' the subject of the equation

F' = mv²/2d.................. Equation 2

From the question,

Given:

  • m = 20 g = 0.02 kg
  • v = 60 m/s
  • d = 40 cm = 0.4 m

Substitute these values into equation 2

  • F' = 0.02(60²)/(0.4×2)
  • F' = 72/0.8
  • F' = 90 N

Hence, The average stopping force acting on the arrow is 90 N

Learn more about average stooping force here: brainly.com/question/13370981

5 0
2 years ago
The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^
allochka39001 [22]

Answer:

3 820 885 N

Explanation:

Gravitational equation

   F = G  m1 m2 / r^2    

         G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11   *   4.41 x 10^5  * 5.97 x 10^24  / ( 6.78x 10^6)^2

 = 3820885 .3 N

6 0
2 years ago
A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k
Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

I_1=m_1r^2

I_1=7\times (0.25)^2=0.437\ kgm^2

The moment of inertia of the rod about one end is given by :

I_2=\dfrac{m_2l^2}{3}

l = r

I_2=\dfrac{m_2r^2}{3}

I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2

For 6 spokes, I_2=0.025\times 6=0.15\ kgm^2

So, the net moment of inertia of the wagon is :

I=I_1+I_2

I=0.437+0.15=0.587\ kgm^2

So, the moment of inertia of the wagon wheel for rotation about its axis is 0.587\ kgm^2. Hence, this is the required solution.

4 0
3 years ago
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