Answer: 1/R = 1/R1 + 1/R2+ ...+ 1/Rn
R is resistance of system in which there are resistors R1, R2 , ... Rn parallel.
<em>Answer:</em>
<em>The answers are: </em>
- <em>A-which is the image is always right side up.</em>
- <em>E-the image is virtual</em>
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<em>Explanation: MY EXPLANATION IS YOU ARE WELCOME BIG DOG 100..</em>
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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
Hence, The average stopping force acting on the arrow is 90 N
Learn more about average stooping force here: brainly.com/question/13370981
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
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 :
The moment of inertia of the rod about one end is given by :
l = r
For 6 spokes, 
So, the net moment of inertia of the wagon is :
So, the moment of inertia of the wagon wheel for rotation about its axis is 
. Hence, this is the required solution.
