Answer:
The transverse component of acceleration is 26.32 where as radial the component of acceleration is 8.77
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s
So the transverse component of acceleration are given as
Here
So
The transverse component of acceleration is 26.32
The radial component is given as
Here
So
The radial component of acceleration is 8.77
Answer:
F_Balance = 46.6 N ,m' = 4,755 kg
Explanation:
In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in
∑ F = 0
Fe –W + F_Balance = 0
F_Balance = - Fe + W
The electric force is given by Coulomb's law
Fe = k q₁ q₂ / r₂
The weight is
W = mg
Let's replace
F_Balance = mg - k q₁q₂ / r₂
Let's reduce the magnitudes to the SI system
q₁ = + 8 μC = +8 10⁻⁶ C
q₂ = - 3 μC = - 3 10⁻⁶ C
r = 0.3 m = 0.3 m
Let's calculate
F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²
F_Balance = 49 - 2,397
F_Balance = 46.6 N
This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.
Mass reading is
m' = F_Balance / g
m' = 46.6 /9.8
m' = 4,755 kg
Answer:
Explanation:
The angular momentum of the pulsar is given by:
where
is the mass of the pulsar
is the radius
is the angular speed
Given the period of the pulsar, , the angular speed is given by
And so, the angular momentum is
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s