the answer is B- are renewable resources
Explanation:
F net of sled = Tension force by rope - Kinetic friction between ground.
F normal of sled = mg = (67kg)(9.81kg/m^2) = 657.27N.
Kinetic friction = 0.18 (I cannot see the value) * Normal force of sled = 0.18 * 657.27N = 118.31N
So F net of sled = 800N - 118.31N = 681.69N.
(I cannot see what the question is asking for, please check on your own!)
Answer:
0.51 m
Explanation:
Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.
Kinetic energy, KE=½kx²
Where k is spring constant and x is the compression of spring
Potential energy, PE=mgh
Where g is acceleration due to gravity, h is height and m is mass
Equating KE=PE
mgh=½kx²
Making x the subject of formula

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

Answer:


Explanation:
k = Coulomb constant = 
Q = Charge
r = Distance = 8 cm
R = Radius = 4 cm
Electric field is given by

Volume charge density is given by

The volume charge density for the sphere is 

The magnitude of the electric field is 