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3 years ago

9

What two factors do u need to know to calculate how much work was done in any situation

2 answers:

pickupchik [31]3 years ago

7 0

U need force and direction to calculate work done.
W= FD

Dafna1 [17]3 years ago

5 0

Force and direction to calculate work done.
<span>W= FD</span>

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3 years ago

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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

5 0

1 year ago

A charge Q = 1.96 10-8 C is surrounded by an equipotential surface with a surface area of 1.18 m2. what is the electric potentia

den301095 [7]

Answer:

V = 575.6 Volts

Explanation:

As we know that surface area of the equi-potential surface is given as

$A = 1.18 m^2$

so we will say

$A = 4\pi r^2$

$1.18 = 4\pi r^2$

$r = 0.31 m$

Now the potential due to a point charge is given as

$V = \frac{kQ}{r}$

$V = \frac{(9\times 10^9)(1.96 \times 10^{-8})}{0.31}$

$V = 575.6 Volts$

8 0

3 years ago