The solution for this problem is:
Let u denote speed.
Equating momentum before and after collision:
= 0.060 * 40 = (1.5 + 0.060) u
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.
Answer:
<h2>Magnitude of the second charge is
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Explanation:
According to columbs law;
F = 
F is the attractive or repulsive force between the charges = 12N
q1 and q2 are the charges
let q1 = - 8.0 x 10^-6 C
q2=?
r is the distance between the charges = 0.050m
k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²
On substituting the given values
12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²
Cross multiplying
Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
Answer:
The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
Explanation:
Electric potential is given as;
V = E*r
where;
E is the electric field strength, = kq/r²
V = ( kq/r²)*r
V = kq/r
k is coulomb's constant = 8.99 X 10⁹ Nm²/C²
q is the charge of the particles = 1.6 X 10⁻¹⁹ C
r is the distance between the particles = 859 nm
At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm
V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)
V = 3.349 X 10⁻³ Volts
Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
Answer:
poor diet andlack of exersise
Explanation:
