Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F = 
q = 
we calculate
q = 
q = 
Ra 7 10-12
q = 2.65 10⁻⁶ C
The Kelvin scale has no negatives on it.
Zero Kelvin is 'Absolute Zero', and nothing can get colder than that.
Answer:
You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.
Answer:
a) 19440 km/h²
b) 10 sec
Explanation:
v₀ = initial velocity of the car = 45 km/h
v = final velocity achieved by the car = 99 km/h
d = distance traveled by the car while accelerating = 0.2 km
a = acceleration of the car
Using the kinematics equation
v² = v₀² + 2 a d
99² = 45² + 2 a (0.2)
a = 19440 km/h²
b)
t = time required to reach the final velocity
Using the kinematics equation
v = v₀ + a t
99 = 45 + (19440) t
t = 0.00278 h
t = 0.00278 x 3600 sec
t = 10 sec
