Answer: 0.03 N/C
Explanation:
We use the current density formula to solve this question.
I/A = σ * E
Where,
I = current flowing in the circuit = 0.3 A
A = cross sectional area of the wire = 1 mm²
σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1
E = strength of the electric field required
I/A = σ *E
E = I/(A * σ)
First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m
E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)
E = 0.3 A / 10 Ω^-1
E = 0.03 N/C
To solve this problem we will apply the concepts related to the kinematic equations of linear emotion. We will start from the definition of acceleration as the change of speed as a function of time. Later we will apply the second law of kinematics for which displacement is understood as a function of initial speed, time and acceleration. We will use the expressions given initially for acceleration as a function of velocity and we will obtain a system of equations that will allow us to find the initial velocity, and subsequently the acceleration
PART A)
Position is given as,
Replacing,
Reorganizing to find the initial speed
Replacing we have,
Therefore the speed of Antelope at first point is 5.23m/s
PART B)
The acceleration of the Antelope would be
Therefore the acceleration of the Antelope is
Answer:
centripetal acceleration
Explanation:
convert 1.63rev/sec to rpm by multiplying by 60
= 1.63*60=97.8rpm
Convert this to rad/sec
1rpm =π/30 rad/sec
97.8rpm = 97.8 * (π/30 rad/sec)
=10.25rad/sec
linear velocity= angular velocity *radius
radius =13.2cm=13.2/100=0.132m
v=rω
v= 0.132*10.25
v=1.35m/s
centripetal acceleration =
The king gave them all fake seeds, the little girl was the only honest child who didn't switch seeds.
Answer:
Explanation:
1 ) Since it is a isochoric process , heat energy passed into gas
= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .
= 7.4 x 12.47 x ( 500 - 300 )
= 18455.6 J.
2 ) Since there is no change in volume , work done by the gas is constant.
3 ) from , gas law equation
PV = nRT
P = nRT / V
= 7.4 x 8.3 x 500 / .74
= .415 x 10⁵ Pa.
4 ) Average kinetic energy of gas molecules after attainment of final temperature
= 3/2 x R/ N x T
= 1.5 x 1.38 x 10⁻²³ x 500
= 1.035 x 10⁻²⁰ J
1/2 m v² = 1.035 x 10⁻²⁰
v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶
= 1.49 x 10⁶
v = 1.22 x 10³ m /s
5 ) In this process , pressure remains constant
gas is cooled from 500 to 300 K
heat will be withdrawn .
heat withdrawn
= n Cp dT
= 7.4 x 20.79 x 200
= 30769.2 J .
6 )
gas will have reduced volume due to cooling
reduced volume = .74 x 300 / 500
= .444 m³
change in volume
= .74 - .444
= .296 m³
work done on the gas
= P x dV
pressure x change in volume
= .415 x 10⁵ x .296
= 12284 J.