This diagram represents a top-down view of an experiment on a table. The 250g and 100g masses are falling and are pulling the bl
ock.
What is the location and minimum number of masses that would have to be added to the picture above to cause the block to move only toward the bottom (straight down)?
A. 1000g mass on bottom
B. 500g mass on bottom
C. 250g mass on the left
D. 250g mass on left and 200g mass on bottom
What is the location and minimum number of masses that would have to be added to the picture above to cause the block to move only toward the bottom (straight down)?
A. 1000g mass on bottom
B. 500g mass on bottom
C. 250g mass on the left
D. 250g mass on left and 200g mass on bottom
2 answers:
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Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate
W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm
Draw a diagram to illustrate the problem as shown below.
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
As we know that electrostatic force between two charges is given as
here we know that electrostatic repulsion force is balanced by the gravitational force between them
so here force of attraction due to gravitation is given as
here we can assume that both will have equal charge of magnitude "q"
now we have
now we have