The accurate answer is 36 miles per hour
Explanation:
To calculate the average speed, start by determining the total distance and total time Lucy traveled, as average speed is the total distance divided by the total time.
Total Distance
Lucy is known to have covered 2/3 of the total distance initially, later finishing the journey with 60 miles. Therefore, the last segment of 60 miles corresponds to 1/3 of the total distance.
(Total distance)
Consequently, since 60 miles equals one-third, multiplying this by 3 gives the total distance: 180 miles.
This indicates the total distance traversed is 180 miles
Total Time
Lucy took 3 hours for the first part of her journey and 2 hours for the second, leading to a total time of 5 hours (3 + 2 = 5)
Determining Average Speed
Answer:
Explanation:
Examining the derived equation reveals
a=gsinθ−μkgcosθ
All components in this formulation are accurate, but further simplification is achievable.
We can explore additional simplification within the equation.
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This makes option a incorrect, as further simplifications yield
a=g(sinθ−μkcosθ)
Option b is valid and represents the optimal choice.
Since we are provided with g=9.8m/s²
we can also input this into option a
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a=9.8m/s²(sinθ−μkcosθ)
Option C also holds but is less preferred as it substitutes the gravity value directly without prior simplification.
The optimal form would be
a=9.8m/s²(sinθ−μkcosθ)
Thus, the best choice is B.
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According to Newton's second law, Force equals the rate of change of momentum over time. Momentum change is equal to Force times time. So, F=ma can be rearranged to a=F/m, a more recognizable formulation of Newton's second law
Using a relevant kinematic equation for mass m: V=u+at; where initial speed u=0; thus, acceleration a=F/m gives V=(F/m)xt, which translates to t=mV/F. For mass 2m, applying the same formula: V=u+at; u=0; a=F/2m indicates V=(F/2m)xt, leading to t=2mV/F (possibly double the initial time)
I might have erred somewhere along the line, but the fundamental concept seems valid... using another kinematic equation for m: s=ut + (1/2)at²; with s=d; and initial speed u=0; a=F/m; t=1; results in d=(1/2)(F/m) = F/2m. Similarly, for 2m: s=ut + (1/2)at²; s=d; u=0; a=F/2m; and t=1 gives d=(1/2)(F/2m)=F/4m (half the distance perhaps???? WHAT???!)
Response:
Magnitude of the electrostatic force acting on the +32 µC charge, 
Clarification:
Let q₁ = +32 µC, located at x₁ = 0
q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m
q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m
Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).
Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).
The resultant electrostatic force on the 32 µC charge is 
Answer:
1/7 kg
Explanation:
Refer to the attached diagram for enhanced clarity regarding the question.
One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.
g denotes the acceleration due to gravity.
Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.
Given M = 1.0 kg and a = 3/4g.
By applying Newton's second law; 
For the body with mass m;
T - mg = ma... (1)
For the body with mass M;
Mg - T = Ma... (2)
Combining equations 1 and 2 gives;
+Mg -mg = ma + Ma
Ma-Mg = -mg-ma
M(a-g) = -m(a+g)
Substituting M = 1.0 kg and a = 3/4g into this equation leads to;
3/4 g-g = -m(3/4 g+g)
3/4 g-g = -m(7/4 g)
-g/4 = -m(7/4 g)
1/4 = 7m/4
Multiplying gives: 28m = 4
m = 1/7 kg
Hence, the mass of the other box is 1/7 kg
