Answer:
40 cm
Explanation:
We are given that
Load=800 N
Effort=200 N
Load distance=10 cm
We have to find the effort distance.
We know that

Using the formula

Effort distance=
Effort distance=
Effort distance=40 cm
Hence, the effort distance will be 40 cm.
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
The magnitude of the adhesive force allows the top block to remain
attached to the bottom when the blocks and the wire are balanced.
The option that must be true is;
Reason:
The tension exerted by the wire attached to the top block = T
Magnitude of the adhesive = 
Weight of the top block = 
Weight of the bottom block = 
Given that with the exertion of the tension, the two blocks remain at rest, we have;
- T =
+ 
The adhesive causes the bottom block to remain attached to the top block, we have;
Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;
The magnitude of the adhesive force = The weight of the bottom block
Therefore;
= 
Learn more here:
brainly.com/question/18907970