Answer:
the final velocity of the two blocks is
the distance that A slides relative to B is
Explanation:
From the diagram below;
acceleration of A relative to B is : 
where
v = u + at

Making t the subject of the formula; we have:


which implies the distance that A slides relative to B.
The final velocities of the two blocks can be determined as follows:
v = u + at

Thus, the final velocity of the two blocks is
Answer
given,
distance = 140 m
time, t = 3.6 s
moving speed = 53 m/s
a) distance = (average velocity) x time


v₀ + 53 = 77.78
v₀ = 24.78 m/s or 25 m/s
b) 

a = 7.8 m/s²
using equation of motion
v₀² = v₁² + 2 a s
53² = 0²+ 2 x 7.8 x s
s = 180 m
The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:

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Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
The electric force between two charge objects is calculated through the Coulomb's law.
F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m.