Answer:
Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is
Now the distance from the charge is
Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.
Answer: Option (a) is the correct answer.
Explanation:
When these two conducting spheres are connected together through a thin wire then charge from the smaller sphere will travel through the wire. And, this charge will continue to travel towards the neutral sphere until the charge on both the spheres will become equal to each other.
For example, charge on small sphere is 5 C then this charge will continue to travel towards the neutral sphere until its charge also becomes equal to 5 C.
Hence, then their potential will also become equal.
Thus, we can conclude that the spheres are connected by a long, thin wire, then after a sufficiently long time the two spheres are at the same potential.
Answer:
magnetic flux ΦB = 0.450324 ×
weber
current I = 1.02484 
A
Explanation:
Given data
length a = 2.2 cm = 0.022 m
width b = 0.80 cm = 0.008 m
Resistance R = 0.40 ohms
current I = 4.7 A
speed v = 3.2 mm/s = 0.0032 m/s
distance r = 1.5 b = 1.5 (0.008) = 0.012
to find out
magnitude of magnetic flux and the current induced
solution
we will find magnitude of magnetic flux thorough this formula that is
ΦB = ( μ I(a) /2 π ) ln [(r + b/2 ) /( r -b/2)]
here μ is 4π × put all value
ΦB = (4π × 4.7 (0.022) /2 π ) ln [(0.012+ 0.008/2 ) /( 0.012 -0.008/2)]
ΦB = 0.450324 × weber
and
current induced is
current = ε / R
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
put all value
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
current = 4π × (4.7) (0.022) (0.008) (0.0032) / 2π(0.40) [(0.012² ) - (0.008/2 )² ]
current = 1.02484 A
From the Newton’s First Law, we can see that acceleration
is simply the ratio of Force over mass. In this case, mass is the sum of the
mass of each car, that is:
mass = 2300 kg + 2500 kg = 4800 kg
So the formula is:
acceleration = Force / mass
acceleration = 18,000 N / 4800 kg
acceleration = 3.75 m/s^2
In 2 significant figures:
<span>acceleration = 3.8 m/s^2</span>
<span>Answer: 1600 J
Explanation:
1) Data:
a) ideal gas: ⇒ pV = nRT and work = ∫ pdV
b) slowly compressed ⇒ constant temperature and not heat exchange
c) pressure: p = 2 atm
d) intitial volume: Vi = 10 liters
e) final volumen: Vf = 2 liters.
f) then the
volume of the gas is held constant ⇒ not work in this stage.
g) calculate the work done on the gas: W = ?
2) Equation
W = ∫pdV
3) Solution:
Since p = constant, W = p ∫dV = p ΔV = p (Vf - Vi)
p = 2 atm × 1.0 ×10⁵ Pa / atm = 200.000 Pa
Vi = 10 liter × 0.001 m³ ./ liter = 0.01 m³
Vf = 2 liter × 0.001 m³ / liter = 0.002 m³
W = 200.000 Pa × (0.002 m³ - 0.01m³) = - 1.600 J.
The negative sign means the work is done over the system.
That is all the work in the system because at the second stage the volume is held constant.
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