Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV =
mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV =
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p =
9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² =
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = 
v/c= 2.33 10⁻⁴
Answer:
50 kg
Explanation:
Given,
Force ( F ) = 100 N
Acceleration ( a ) = 2 m/s^2
To find : Mass ( m ) = ?
Formula : -
F = ma
m = F / a
= 100 / 2
m = 50 kg
Therefore, the mass of the object is 50 kg.
The 'net' force acting on the box is (9 - 3) = 6 newtons
in the direction that Carlos is pushing.
Force = (mass) x (acceleration)
6 = (3) x (acceleration)
Divide each side by 3 :
<em>2 m/s² = acceleration</em>
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