Given:
B(Magnetic field): 1.5 T
q= 7.5 microcoulombs
v= 1.75 x 10 ∧6 m/s
The angle ∅ between B and v is 45 °.
Now we know that F= qvB sin ∅
Substituting these values we get:
F= 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45
F= 16.752 N
Answer:
Explanation:
a ) F = (-kx + kx³/a²)
intensity of field
I = F / m
= (-kx + kx³/a²) / m
If U be potential function
- dU / dx = (-kx + kx³/a²) / m
U(x) = ∫ (kx - kx³/a²) / m dx
= k/m ( x²/2 - x⁴/4a²)
b )
For equilibrium points , U is either maximum or minimum .
dU / dx = x - 4x³/4a² = 0
x = ± a.
dU / dx = x - x³/a²
Again differentiating
d²U / dx² = 1 - 3x² / a²
Put the value of x = ± a.
we get
d²U / dx² = -2 ( negative )
So at x = ± a , potential energy U is maximum.
c )
U = k/m ( x²/2 - x⁴/4a²)
When x =0 , U = 0
When x= ± a.
U is maximum
So the shape of the U-x curve is like a bowl centered at x = 0
d ) Maximum potential energy
put x = a or -a in
U(max) = k/m ( x²/2 - x⁴/4a²)
= k/m ( a² / 2 - a⁴/4a²)
= k/m ( a² / 2 - a²/4)
a²k / 4m
This is the maximum total energy where kinetic energy is zero.
Value of resistor = (12V) /(1.2 x 10^-3A)=10000ohms=10k ohms
Answer:
n=0.03928 moles
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
Explanation:
The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs
Volume of oxygen in lungs =V=5*20%= 1 L=
Temperature=T=
Pressure at sea level = P= 1 atm=
R is universal Gas Constant =8.314 J/mol.K
Formula:
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
Answer:
Explanation:
It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.
Let in equilibrium , tension in rope be T
For balancing Joe
T = M g
For balancing Simon
friction + T = mgsinθ
μmgcosθ+T = mgsinθ
μmgcosθ+Mg = mgsinθ
M = (msinθ - μmcosθ)
M = m(sinθ - μcosθ)
