(4) a metal sphere with a charge of 1.0 × 10^−9 C <span>moved through a potential difference of 4.0 V would undergo the greatest change in electrical energy from the list. </span>
Directly outside of the nucleus you have a membrane
Compared to the charge on a proton, the amount of charge on an electron is same and has the opposite sign
Answer:
The car overtakes the truck at a distance d = 3266.2ft from the starting point
Explanation:
Problem Analysis
When car catches truck:
dc = dt = d
dc: car displacement
dt: truck displacement
tc = tt = t
tc: car time
tt : truck time
car kinematics :
car moves with uniformly accelerated movement:
d = vi*t + (1/2)a*t²
vi = 0 : initial speed
d = (1/2)*a*t² Equation (1)
Truck kinematics:
Truck moves with constant speed:
d = v*t Equation (2)
Data
We know that the acceleration of the car is 3.00 ft / s² and the speed of the truck is 70.0 ft / s .
Development problem
Since the distance traveled by the car is equal to the distance traveled by the truck and the time elapsed is the same for both, then we equate equations (1 ) and (2)
Equation (1) = Equation (2)
(1/2)*a*t² = v*t
(1/2)*3*t² = 70*t (We divide both sides by t)
1.5*t = 70
t = 70 ÷ 1.5
t = 46.66 s
We replace t = 46.66 s in equation (2) to calculate d:
d = 70*46.66 = 3266.2ft
d = 3266.2 ft
Answer:
The acceleration of the crate is .
Explanation:
Given that,
Force, F = 750 N
Mass of the crate, m = 250 kg
The coefficient of friction is 0.12.
We need to find the acceleration of the crate. The net force acting on the crate is given by :
f is frictional force,
So, the acceleration of the crate is . Hence, this is the required solution.