Answer:
t = o.6 s
Explanation:
Let ball thrown from below be A and ball dropped from above be B.
A and B meet when they both are same level above the ground. Then let A moved up a distance d and B dropped a distance h. Then you know
d + h = 15 m ---------------(1)
Now apply s = ut + 
at²
To A upwards,
d = 25t - gt² -----------------(2)
To B downwards,
h = 0 + gt² ----------------(3)
(1) = (2) + (3) ⇒ 15 = 25t
t = 0.6 s
Answer:
what is the question please?
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:
The value of the final velocity is 
Explanation:
Generally from kinematic equation
Here a is the acceleration which is equivalent to g
So
substituting 6 m for s , 4.2m/s for u
