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3 years ago

What is the IMA of the following pulley system?

Physics

2 answers:

tamaranim1 [39]3 years ago

8 0

Answer ideal mechanical advantage

Irina-Kira [14]3 years ago

7 0

The full form of IMA is an “ideal mechanical advantage”

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According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energ

Lera25 [3.4K]

Answer:

c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R

V2 / V1 = (R1 / R2) = 1/4

V2 = V1 / 4

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3 years ago

Considering this analogy (current = rate that heat flows through a window, and voltage = temperature difference across the windo

erastovalidia [21]

Answer:

Explained

Explanation:

Resistance R in a current flow through an object is given by

$R= \frac{\rho L}{A}$

ρ = resistivity of the material

L= length of the object

A= area of cross section

clearly resistance is directly dependent on length of the object.This means greater the length larger will be resistance to current.

thermal resistance R_th is given by

$R_{th} = \frac{L}{KA}$

L= length of the object

A= area of cross section

K = Conductivity of the material

thermal resistance is also is directly dependent on length of the object.This means greater the length larger will be resistance to current.

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3 years ago

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Lesechka [4]

Answer:

if you only have to control your chakra and know how to get all your vibes to pass it to objects and it takes time to practice

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3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

The picture shows an object resting on a balance.

Maslowich

Answer:

4.90kgm^-2

Explanation:

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2 years ago