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2 years ago

Describe one common critique or criticism of Utilitarianism. To what extent do you agree or disagree with this criticism?

1 answer:

lisabon 2012 [21]2 years ago

7 0

The second most common criticism of utilitarianism is that it is impossible to apply - that happiness (etc) cannot be quantified or measured, that there is no way of calculating a trade-off between intensity and extent, or intensity and probability (etc), or comparing happiness to suffering.

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What is the smallest particle of the element iron (fe) that can be as iron ?

grin007 [14]

This answer is the same for all elements, a single atom of any element is still that element.

6 0

3 years ago

What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movem

Step2247 [10]

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

= 1/5 ∫ 3*10^-3 d(t) + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

The reading of the voltmeter ( V )

V = IR

= 0.96 mA * 21 k Ω = 20.16 v

3 0

2 years ago

A circular wooden loop of mass m and radius R rest on a flat horizontal friction less surface. A bullet, also of mass m, and mov

bearhunter [10]

Answer:

w = vR/3

Explanation:

The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.

From the principle of conservation of linear momentum , we have

m*v = 2*m* Vcm

Where v = velocity of bullet, Vcm = velocity of wood

Hence, we have

Vcm = v2

Also, from the conservation of angular momentum about the centre of mass.

M*V*(R/2) = Ic*w - equation (I)

where Ic = moment of inertia and w = angular velocity

Ic for a ring is given by $mr^2 + m(r/2)^2$

Ic of a bullet is given by $m(r/2)^2$

Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives

Ic(system) = $3*m*R^2/2$

Substituting back into equation (I), we have

$m*v*R^2=3*m*R^2*w/2$

Hence, we obtain w =vR/3

w=v3R

4 0

2 years ago

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0

3 years ago

A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the ground. How far, in meters, from the deck doe

Vesna [10]

Answer:

2.605m

Explanation:

Using the formula for calculating Range (distance travelled in horizontal direction)

Range R = U√2H/g

U is the speed = 4.8m/s

H is the maximum height = ?

g is the acc due to gravity = 9.8m/s²

R = 3.5m

Substitute into the formula and get H

3.5 = 4.8√2H/9.8

3.5/4.8 = √2H/9.8

0.7292 = √2H/9.8

square both sides

0.7292² = 2H/9.8

2H = 0.7292² * 9.8

2H = 5.21

H = 5.21/2

H = 2.605m

Hence the height of the ball from the ground is 2.605m

7 0

2 years ago