Answer:
Explanation:
This problem is all about torque. The "rules" are that in order for a system to be in rotational equilibrium, the sum of the torques on the system have to equal 0 (in other words, they have to equal each other {cancel each other out}). The equation for torque is
τ = F⊥r where τ is torque, F⊥ is the perpendicular force, and r is the lever arm length in meters. We also have to understand that in general Forces moving clockwise are negative and Forces moving counterclockwise are positive. Now we're ready for the problem:
A. The counterclockwise torque:
τ = 300(3) so
τ = 900N*m
B. The clockwise torque:
τ = -450(2.5) so
τ = -1100N*m
C. Obviously the system is not in roational equilibrium because one side is experiencing a greater torque than the other. This system will move clockwise as it currently exists.
D. In order for the system to be in rotational equilibrium, we have to move Bob's location from the fulcrum. Let's see to where.
The torques have to be the same on both sides of the fulcrum; mathematically, that looks like this:
F⊥r = F⊥r Filling in:
300(3) = 450r and
900 = 450r so
2 = r. This means that Bob will have to move closer to the fulcrum by a half of a meter to 2 meters from the fulcrum in order for the system to be in balance.
Isn't this so much fun?!
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance
so Work done = 0.085 x 20g x 28
= 466.956 J
Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ = 
= 
= 3.61cm = 0.036m
r₂ = 
= 
= 5cm = 0.05m
electric potential V = 
change in potential ΔV = 
ΔV = 
, where 
2.00μC
ΔV = 
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × 
ΔV= 2.789×10⁵
= ΔV × q₃
ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s
Answer:
Heat required to melt 1 lb of ice is 151.469 KJ
Explanation:
We have given mass of ice = 1 lb
We know that 1 lb = 0.4535 kg
Latent heat of fusion for ice =334 KJ/kg
Amount if heat for fusion of ice is given by
, here m is mass of ice and L is latent heat of fusion
So heat 
So heat required to melt 1 lb of ice is equal to 151.469 KJ
Answer:
The answer to your question is : 521.8 m
Explanation:
Data:
Different heights
Time first object (tfo) = 10.7 s
Time second object (tso)= 14.8 s
Initial speed of both objects(vo) = 0 m/s
a = 9.81 m/s²
Formula:
h = vot + 1/2 (a)(t)² but vo = 0 so, h = 1/2 (a)(t)²
Then, height fo h = 1/2 (9.81)(10.7)² = 561.6 m
height so h = 1/2(9,81)(14.8)² = 1074.4 m
Difference in their heights = 1074.4 m - 561.6 m = 521.8 m
