Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B

Where
=Mass of block B
Substitute the values


For block A

Where
Mass of block A
Substitute the values


Hence, the friction force acting on block A=40 N
Answer: Given:
Initial velocity= 36km/h=36x5/18=10m/s
Final velocity =54km/h=54x5/18=15m/s
Time =10sec
Acceleration = v-u/ t
=15-10/10=5/10=1/2=0.5 m/s2
Distance =s=?
From second equation of motion:
S=ut +1/2 at^2
=10*10+1/2*0.5*10*10
=100+25
=125m
So distance travelled 125m
Hope it helps you
Answer:
- the expected value is 8
- the standard deviation is 2.8284
Explanation:
Given the data in the question;
The model N(t), the number of planets found up to time t, as a poisson process,
∴ N(t) has distribution of poisson distribution with parameter (λt)
so
the mean is;
λ = 1 every month = 1/3 per month
E[N(t)] = λt
E[N(t)] = (1/3)(24)
E[N(t)] = 8
Therefore, the expected value is 8
For poisson process, Variance and mean are the same,
Var[N(t)] = Var[N(24)]
Var[N(t)] = E[N(24)]
Var[N(t)] = 8
so the standard deviation will be;
σ[N(24)] = √(Var[N(t)] )
σ[N(24)] = √(8 )
σ[N(24)] = 2.8284
Therefore, the standard deviation is 2.8284


Now


- Lower mass=Higher acceleration
- Lower Force=Lower Acceleration
Option B has lowest mass and highest force hence its correct
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I