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2 years ago

How much work does it take to move a 50 μC charge against a 12 V potential difference?

1 answer:

8 0

<span>work =V*Q =12*50*10^-6

The total work done will be equal to

work = V.Q

which means

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>

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In which condition is relative velocity known by adding the velocity od the first body and that of the second body ?

Softa [21]

Answer:

when a two bodies A and B are moving at an angle 180° with each other then the relative velocity is the sum of bodies the velocity .i.e,

when the bodies move the opposite direction

then their relative velocity is the sum of individual velocities.

3 0

2 years ago

A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string

krek1111 [17]

Angular velocity = (75x2pie)/60
      =2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
       v= 0.5 x 2.5pie
   v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
      F= (0.2 x 3.93^2)/0.5
      F=6.18 N
(sorry if wrong)

3 0

2 years ago

Read 2 more answers

Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$