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frozen [14]
3 years ago
15

I love making friends but i always feel lonely around them, i try to act fine and as if nothing bothers me but it does. I want a

steady partner who loves me for me. Is it to much to ask for? People are mad because I'm clingy which is something I'm working on but I don't like people hating me for little things I do. I try to make friends but everyone makes it hard for me. I'm sorry I made you read this.. but sometimes talking about these things help..
Physics
2 answers:
arsen [322]3 years ago
6 0

Answer:

i will be your friend ...

Ivan3 years ago
4 0

Answer:

Explanation:

its good you let it out, just remember youre not alone. just keep trying to improve on the things you want to improve on and focus on yourself. if you feel lonely around the friends you made, then they might not be good for you, i know its hard to let go because they are the only people you have but trust me, its better if you let go of them and take time to really get to know yourself before getting to know anyone else.

before you get a steady partner, learn to love yourself. it sound like you didnt  but are trying now which is great! keep it going because before you can love ANYBODY you must love yourself. Keep fighting and i'll be rooting for you!!

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A 7.5 cm object is 14 cm from a concave lens, which has a focal length of –7 cm. Its image is 5 cm in front of the lens. What is
Rama09 [41]
Given conditions:
height of object = 7.5cmdistance of object from mirror  = 14 cmfocus length = -7 cmimage distance =  ?
Using mirror formula: 
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)


Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
3 0
3 years ago
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Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
larisa86 [58]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • Mass of the larger disk = M_2\ =\ 1.60\ kg
  • Mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the larger disk = R_2\ =\ 5.00\ cm\ =\ 0.05\ m
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
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Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)

From the pulley,

\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)

From the equation (1) and (2),

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = R_2\ =\ 0.05\ m

Let a_2 be the acceleration of the block while connecting to the larger pulley.\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2

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The table shows the potential energy and kinetic energy of a skier at two different positions on a hill.
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Answer:

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Nutka1998 [239]

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<h3>Which device uses electromagnetic induction to transfer electrical energy from one circuit to another?</h3>

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To learn more about transformers refer:

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