Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s
Answer:
A)
0.395 m
B)
2.4 m/s
Explanation:
A)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= initial position of spring from equilibrium position = 0.21 m
= initial speed of the cart = 2.0 ms⁻¹
= amplitude of the oscillation = ?
Using conservation of energy
Final spring energy = initial kinetic energy + initial spring energy
B)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= amplitude of the oscillation = 0.395 m
= maximum speed at the equilibrium position
Using conservation of energy
Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring
The new volume = 3 x 52.6 that’s because as the pressure decreases by 1/3 the volume increases x3
The answer is a property of density. The higher the density, the higher the pressure at the bottom.
Pressure = mass / Area. So given that the 4 samples occupy the same area at the bottom, the mass is going to be the determining factor. Per given volume, mercury has the largest mass. The answer is A
1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:
where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find
2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force
where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
