Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
F = ma
where force is magnetic force
F = q v x B
the bold are vectors, if we write the module of this expression we have
F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
F = q vB
the acceleration of the particle is centripetal
a = v² / r
we substitute
qvB = m v² / r
qBr = m v
q =
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
v = d / t
the distance is ¼ of the circle,
d =
d =
we substitute
v =
r =
let's calculate
r =
2 2.2 10-3 88 /πpi
r = 123.25 m
let's substitute the values
q =
7.2 10-8 88 / 0.6 123.25
q = 8.57 10⁻⁸ C
Let's reduce to mC
q = 8.57 10⁻⁸ C (10³ mC / 1C)
q = 8.57 10⁻⁵ mC
Answer:
0.37 m/s to the left
Explanation:
Momentum is conserved. Initial momentum = final momentum.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
Initially, both the fisherman/boat and the package are at rest.
0 = m₁ v₁ + m₂ v₂
Plugging in values and solving:
0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)
v = -0.37 m/s
The boat's velocity is 0.37 m/s to the left.
Answer:
17 °C
Explanation:
From specific Heat capacity.
Q = cm(t₂-t₁)................. Equation 1
Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.
make t₁ the subject of the equation
t₁ = t₂-(Q/cm)............... Equation 2
Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c
Substitute into equation 2
t₁ = 22-[5000/(4×250)
t₁ = 22-(5000/1000)
t₁ = 22-5
t₁ = 17 °C
Point a because point a is the highest at potential energy converting into the highest kinetic energy.