Answer:
A.) 8 m/s
B.) 7.0 m
Explanation:
Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.
(a) What is its velocity when it reaches the top of the plane?
Since the plane is frictionless, the final velocity V will be the same as 8 m/s
The velocity will be 8 m/s as it reaches the top of the plane.
(b) How far horizontally does it land after it leaves the plane?
For frictionless plane,
a = gsinø
Acceleration a = 9.8sin28
Acceleration a = 4.6 m/s^2
Using the third equation of motion
V^2 = U^2 - 2as
Substitute the a and the U into the equation. Where V = 0
0 = 8^2 - 2 × 4.6 × S
9.2S = 64
S = 64/9.2
S = 6.956 m
S = 7.0 m
Answer:
Covalent Bond is found between the atoms of a molecule.
Answer:
a) q_1=q_2= 7.42*10^-7 C
b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C
Explanation:
Given:
F_e = 0.220 N
separation between spheres r = 0.15 m
Electrostatic constant k = 8.99*10^9
Find: charge on each sphere
part a
q_1 = q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_1^2 = F_e*r^2/k
q_1=q_2= sqrt (F_e*r^2/k)
Plug in the values and evaluate:
q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)
q_1=q_2= 7.42*10^-7 C
part b
q_1 = 4*q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_2^2 = F_e*r^2/4*k
q_2= sqrt (F_e*r^2/4*k)
Plug in the values and evaluate:
q_2= sqrt (0.22*0.15^2/4*8.99*10^9)
q_2= 3.7102*10^-7 C
q_1 = 14.8*10^-7 C
Ummm I’m not sure let me do the work
Voltage is the difference in charge between two points.
Current is the rate the charge flows
Resistance is the tendency a material has to resist the flow of charge (current)
Combining voltage resistance and current Ohm developed the formula
V (Voltage)= I (Current) x R (Resistance)
