Answer:
B = 1.353 x 10⁻³ T
Explanation:
The Magnetic field within a toroid is given by
B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.
To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,
r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.
b = 25.1 + 3 = 28.1 cm
a = 25.1 cm
r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m
B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266
B = 1.353 x 10⁻³ T
The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T
 
        
             
        
        
        
Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) =  7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is

V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
 
        
        
        
Answer:
500 kg
Explanation:
It is given that,
The mass of a open train car, M = 5000 kg
Speed of open train car, V = 22 m/s
A few minutes later, the car’s speed is 20 m/s
We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :
initial momentum = final momentum
Let m is final mass
MV=mv

Water collected = After mass of train - before mass of train
= 5500 - 5000
= 500 kg
So, 500 kg of water has collected in the car.
 
        
             
        
        
        
Answer:
Explanation:
24 - gauge wire , diameter = .51 mm . 
Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m
R = ρ l / s 
1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]
= 8.42 x 10⁻² ohm 
= .084 ohm 
B )  Current required through this wire
= 12 / .084 A
= 142.85 A
C ) 
Let required length be l 
resistance = .084 l 
2 = 12 / .084 l 
l = 12 / (2 x .084)
= 71.42 m 
 
        
                    
             
        
        
        
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges. 
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis. 
force on it due to rest of the charges will be equal and opposite so 
k3q Q / x² =k 8q Q / (L+x)² 
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
 x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal 
kQ . 3q / x² = k3q  8q / L²
Q = 8q  (x² / L²)
so charge required = - 8q  (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )