A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with
1 answer:
Answer:
11.72 mm
Explanation:
The gravitational potential energy equals the potential energy of the spring hence
where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring
where \theta is the angle of inclination and d is the sliding distance
Making x the subject then
Substituting the given values then
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Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = v²
v = √( 2K / )
lets relate the cross-sectional area A of the beam to its diameter D;
A = πD²
now, we substitute for v and A
n = I / πeD² ×√( 2K /
)
n = 4I/π eD² × √( / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³