Answer:
The mass of the cargo is 
Explanation:
From the question we are told that
The radius of the spherical balloon is 
The mass of the balloon is
The volume of the spherical balloon is mathematically represented as

substituting values


The total mass the balloon can lift is mathematically represented as

where
is the density of helium with a value of

and
is the density of air with a value of

substituting values


Now the mass of the cargo is mathematically evaluated as


here we can say that there is no external force on fisherman and dock
so here we will use momentum conservation theory
As per momentum conservation
initial momentum of fisherman + boat = final momentum of fisherman + boat

now we will have



so the speed of boat and fisherman will be 1.16 m/s
Answer
given,
time = 10 s
ship's speed = 5 Km/h
F = m a
a is the acceleration and m is mass.
In the first case
F₁=m x a₁
where a₁ = difference in velocity / time
F₁ is constant acceleration is also a constant.
Δv₁ = 5 x 0.278
Δv₁ = 1.39 m/s

a₁ = 0.139 m/s²
F₂ =m x a₂
F₃ = F₂ + F₁
Δv₃ = 19 x 0.278
Δv₃ = 5.282 m/s
a₃=Δv₂ / t

a₃ = 0.5282 m²/s
m a₃=m a₁ + m a₂
a₃ = a₂ + a₁
0.5282 = a₂ + 0.139
a₂=0.3892 m²/s
F₂ = m x 0.3892...........(1)
F₁ = m x 0.139...............(2)
F₂/F₁
ratio = 
ratio = 2.8
Answer:
The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²
Explanation:
Data
Ta = 0.125 s
Tb = 0.08 s
Δtab = 0.5 s
distance = 1 cm
Process
1.- Calculate va
va = 1/0.125 = 8 cm/s
vb = 1/0.08 = 12.5 cm/s
2.- Calculate Δv
Δv = 12.5 - 8
Δv = 4.5 cm/s
3.- Calculate acceleration
a = Δv / Δt
a = 4.5/0.5
a = 9 cm/s²
Answer:

Explanation:
Given that,
The magnitude of magnetic field, B = 0.55 T
The radus of the loop, r = 43 cm = 0.43 m
The current in the loop, I = 5.8 mA = 0.0058 A
We need to find the magnetic moment of the loop. It is given by the relation as follows :

Put all the values,

So, the magnetic moment of the loop is equal to
.