M1 descending
−m1g + T = m1a
m2 ascending
m2g − T = m2a
this gives :
(m2 − m1)g = (m1 + m2)a
a =
(m2 − m1)g/m1 + m2
= (5.60 − 2)/(2 + 5.60) x 9.81
= = 4.65m/s^2
By wave particle duality.
Wavelength , λ = h / mv
where h = Planck's constant = 6.63 * 10⁻³⁴ Js, m = mass in kg, v = velocity in m/s.
m = 1kg, v = 4.5 m/s
λ = h / mv
λ = (6.63 * 10⁻³⁴) /(1*4.5)
λ ≈ 1.473 * 10⁻³⁴ m
Option D.
Let
upthrust = T
weight = W = mg
Air resistance = F
When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)
Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)
Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
Answer: SG = 2.67
Specific gravity of the sand is 2.67
Explanation:
Specific gravity = density of material/density of water
Given;
Mass of sand m = 100g
Volume of sand = volume of water displaced
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Therefore, the specific gravity of sand is
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
Specific gravity of the sand is 2.67
