1) In a circular motion, the angular displacement
is given by
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire:
. Using the information about the radius,
, we find the total angular displacement:
2) If we put larger tires, with radius
, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
-- Class I lever
The fulcrum is between the effort and the load.
The Mechanical Advantage can be anything, more or less than 1 .
Example: a see-saw
-- Class II lever
The load is between the fulcrum and the effort.
The Mechanical Advantage is always greater than 1 .
Example: a nut-cracker, a garlic press
-- Class III lever
The effort is between the fulcrum and the load.
The Mechanical Advantage is always less than 1 .
I can't think of an example right now.
Answer:
547 m
Explanation:
From law of motion
s = ut + ½at²
Where "t" is Time taken to reach Earth
s= distance= 182 m
a= vertical acceleration = 5.82 m / s 2
U= initial velocity in vertical position = 0
182= ½ × 5.82t²
t²=( 2× 182)/ 5.82
= 364/5.82
= 62.54
t= √62.54
t= 7.908s
horizontal distance travelled = speed x time
Horizontal speed= 72.6 m / s
horizontal distance travelled =72.6× 7.908
= 547 m
Hence, the survivor will it hit the waves at 547 m away
Answer:
1keff=1k1+1k2
see further explanation
Explanation:for clarification
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?
From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.
Now the spring is in series combination
F
e
F=ke
k=f/e.........*
where k is the force constant or the constant of proportionality
k=f/e
............................1
also for effective force constant
divide all through by extension
1) Total force is
Ft=F1+F2
Ft=k1e1+k2e2
F = k(e1+e2) 2)
Since force on the 2 springs is the same, so
k1e1=k2e2
e1=F/k1 and e2=F/k2,
and e1+e2=F/keq
Substituting e1 and e2, you get
1/keq=1/k1+1/k2
Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.
I think the answer would be 70
