The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2
1 answer:
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Answer:
a= 1.59 m/s² : Magnitude of the acceleration
β = 65.22° (north of east) : Direction of the acceleration
Explanation:
Conceptual analysis
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Problem development
The acceleration is presented in the direction of the resultant force applied .
Calculation of the resultant forces (R)
R= 429.5 N
We apply the formula (1) to calculate the magnitude of the acceleration(a) :
∑F = m*a , m= 270 kg
R= m*a
429.5 =270*a
a= 1.59 m/s²
Calculation of the direction of the acceleration (β)
β = 65.22° (north of east)
The magnetic force acting on a charged particle moving perpendicular to the field is:
= qvB
is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.
The centripetal force acting on a particle moving in a circular path is:
= mv²/r
is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.
If the magnetic force is acting as the centripetal force, set equal to
and solve for B:
qvB = mv²/r
B = mv/(qr)
Given values:
m = 1.67×10⁻²⁷kg (proton mass)
v = 7.50×10⁷m/s
q = 1.60×10⁻¹⁹C (proton charge)
r = 0.800m
Plug these values in and solve for B:
B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)
B = 0.979T