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belka [17]
3 years ago
9

The physiological need include:

Physics
1 answer:
harina [27]3 years ago
5 0

Answer:

A

Explanation:

food for energy, security from theft and privacy

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Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

8 0
3 years ago
A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle abov
Digiron [165]

Ox:vₓ=v₀

x=v₀t

Oy:y=h-gt²/2

|vy|=gt

tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g

y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°

cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s

Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ

3 0
3 years ago
A ball of radius 0.220 m rolls along a horizontal table top with a constant linear speed of 3.70 m/s.?
4vir4ik [10]
<span>16.82 x 0.04 = 0.67 rad
I hope I helped if you really need I can explain to you how I got that answer but Thats correct im sorry it took 2 days for me to find this answer but if you or anybody else still needs the answer for this question here it is :) have a fantastic day guys Spring Break is coming up soon :)</span>
6 0
3 years ago
An ax is an example of a ____________.
natta225 [31]
<span>An ax is an example of a wedge. The correct option among all the options that are given in the question is the second option or option "b". The other choices given in the question are incorrect and can be easily neglected. I hope that this is the answer that has actually come to your great help.</span>
4 0
3 years ago
Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

 m_1 = m_2

\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

5 0
3 years ago
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