The force of gravity F_g will act downwards.
Normal force F_N will act upwards equal to the force of gravity.
A force due to uniform acceleration F_a will act upwards to move the elevator upwards.
Thus, figure E is the correct answer.
This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.
I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.
Let the mass of Mass of Object 5 be 20 clods .
Then . . .
-- The mass of Object 2 is double the mass of Object 5 = 40 clods.
-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.
So now, here are the masses:
Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
Object #3 . . . . . 5 clods
Object #4 . . . . . 10 clods
Object #5 . . . . . 20 clods .
Now let's check out the statements, and see how they stack up:
Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
exert the same force on the same mass.
Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
exert the same force on the same mass.
Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).
Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).
Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
the only one that can be true.
Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)
Answer:
a
The hiker (you ) is 200 m below his/her(your) starting point
b
The resultant displacement in the north east direction is
The resultant displacement in vertical direction (i.e the altitude change )
Explanation:
From the question we are told that
The displacement in the morning is 
The displacement in the afternoon is 
Generally the direction west is negative , the direction east is positive
the direction south is negative , the direction north is positive
resultant displacement is mathematically evaluated as
From the above calculation we see that at the end of the hiking the hiker (you) is 200 m below his/her(your) initial position
Generally from Pythagoras theorem , the resultant displacement in the north east direction is
=>
Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )
=>
