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Inessa [10]
3 years ago
11

A plastic rod is rubbed with a piece of wool, and a glass rod is rubbed with a piece of silk. An object is placed near the plast

ic rod, and an attractive force is detected. How will the object react when near the glass rod?
.A repulsive force will occur
.B.An attractive force will occur
.C.A neutralization will occur
.D.An electric discharge will occur
Physics
1 answer:
goldenfox [79]3 years ago
5 0

Answer:

B

Explanation:

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Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m
ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

 a = 62.5²/(2 × 22)

 a = 88.78m/s²

the time you need to get this speed

     v = v₀ + a t

     t = v / a

     t = 62.5 / 88.78

     t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

     t = 1,025 s

      a = 55.43 m / s²

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3 years ago
Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 1.4×109 electrons from one disk to the
allsm [11]

Answer:

r = 6.5*10^-3 m

Explanation:

I'm assuming you meant to ask the diameters of the disk, if so, here's it

Given

Quantity of charge on electron, Q = 1.4*10^9

Electric field strength, e = 1.9*10^5

q = Q * 1.6*10^-19

q = 2.24*10^-10

E = q/ε(0)A, making A the subject of formula, we have

A = q / [E * ε(0)], where

ε(0) = 8.85*10^-12

A = 2.24*10^-10 / (1.9*10^5 * 8.85*10^-12)

A = 2.24*10^-10 / 1.6815*10^-6

A = 1.33*10^-4 m²

Remember A = πr²

1.33*10^-4 = 3.142 * r²

r² = 1.33*10^-4 / 3.142

r² = 4.23*10^-5

r = 6.5*10^-3 m

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3 years ago
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In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t
Pie

Answer:

A) Q_a=74256\ J

B) Q=93562560\ J

Explanation:

Given:

  • temperature of air, T_a=-19+273=254\ K
  • temperature of lungs, T_l=37+273=310\ K
  • specific Heat exchanged from the lungs , c_l=0.47\ J.kg^{-1}.K^{-1}
  • specific heat of air, c_a=1020\ J.kg^{-1}.K^{-1}
  • mass of 1 L air, m'=1.3\ kg
  • breath rate, b=21\ breath.min^{-1}

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

Q_a=m'.c_a.\Delta T

Q_a=1.3\times1020\times (310-254)

Q_a=74256\ J

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

B=b.60

B=21\times 60

B=1260

<u>Now the total loss of energy in 1 hr.:</u>

Q=Q_a.B

Q=74256\times 1260

Q=93562560\ J

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