Answer:
v = 2.33 103 m / s
Explanation:
The magnetic force is
F = q v x B = i L X B
Where L is wire length and B the magnetic field
Let's calculate the value of the force
F = i L B
Now we can calculate the acceleration using Newton's second law
F = m a
a = F / m
Density is
ρ = m / V
V = π r² L
ρ = m / π r² L
m = ρ π r² L
a = I L B / (ρ π r² L)
a = 10 10³ 2.0 / (8960 π (5.1 10⁻⁴)²)
a = 2.73 10⁶ m/s²
Having the acceleration we use the kinematic equation
v² = v₀² + 2 a x
v = √ 2 ax
v = √ (2 2,723 10⁶ 1.0)
v = 2.33 103 m / s
Answer:
570 J
Explanation:
im sorry im not sure about the rest but i hope this helps
Explanation:
A point on the rim of the wheel went from rest to 15 m/s so we can solve for the acceleration as
or
We also know that the angular acceleration is
Using r = 0.30 cm and a = 1.9 m/s^2, we get
Answer:
μ = 0.604
Explanation:
For the cat to stay in place on the merry go round, the maximum static frictional force must be equal in magnitude to that of the centripetal force.
Now, Centripetal force is given as;
Fc = mv²/r
Where r is radius and v is tangential speed and m is mass.
We also know that maximum static frictional force is given by;
F_static = μmg
Where μ is coefficient of friction
Now, equating both forces, we have;
mv²/r = μmg
Divide through by m;
v²/r = μg
Now, tangential speed can be expressed as;
v = circumference/period
Thus, v = 2πr/T
Where T is period of rotation and
2πr is the circumference of the merry go round.
Thus,
v²/r = μg is now;
(2πr/T)²/r = μg
Making μ the subject, we have;
(2πr/T)²/rg = μ
μ = [(2π x 5.4)/6]²/(5.4 x 9.8)
μ = 0.604