Answer:
the force of attraction between the two charges is 4.2 x 10⁹ N.
Explanation:
Given;
the magnitude of first charge, q₁ = 0.06 C
the magnitude of the second charge, q₂ = 0.07 C
distance between the two charges, r = 3 m
The force of attraction between the two charges is calculated as ;
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/C²
Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.
Answer:
the charged particle is electron..
Explanation:
electrons are the only charged particles which can be transferred by rubbing certain materials which generate electrostatic force ...
Answer:
25.0 m / 10 m/s = 2.5 s
50.0 / 9.5 = 5.26 s
25.0 / 11.1 = 2.25 s
T (Wood) = 2.5 + 5.26 + 2.25 = 10.0 s
Mrs Wood runs 10 s vs 10.4 for Mr Overstreet
Answer:
C) 6 m/s
Explanation:
Given that
m₁=5000 kg
The initial velocity of 5000 kg car =u₁
m₂=10,000 kg
The initial velocity of 10000 kg car =u₂ = 0 m/s
After collision the final speed of the both car,v = 2 m/s
There is no any external force on the system that is why linear momentum will be conserved.
Linear momentum P = m v
m₁u₁ + m₂u₂ = (m₂ + m₁) v
5000 x u₁ + 10000 x 0 = (5000 + 10000) x 2
5000 x u₁ = 15000 x 2
5 x u₁ = 15 x 2
u₁ = 6 m/s
Therefore the answer is C.
C) 6 m/s
Convert the given in SI units.
(44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec
The distance traveled and the initial velocity can be related through the equation,
d = (Vf)² - (Vi)²/ 2a
where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,
d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)
The value of d from the equation,
d = 9.17 meters
Convert this to feet,
d = (9.17 m)(3.28 ft / 1 m) = 30 ft
Answer: 30 ft
