Answer:
4375 N, 7875 N
Explanation:
since the body is equilibrium
total upward force = total downward force
w weight of the bridge = FL + FR
when the car was introduced,
total downward force = total upward force
FL₁ + FR₁ = w + (m × acceleration due to gravity) with w = FL + FR
then FL₁ + FR₁ = FL + FR + Mcg
FL₁ + FR₁ - FL - FR = Mcg
ΔFL + ΔFR = 1250 × 9.8 = 12250 N
taken the left as the pivot point and using the principle of moment
ΔFR × 14 m = 12250 N × 5 m + (ΔFL × 0 m) since the left is the pivot point.
ΔFR = 61250 / 14 = 4375 N
but
ΔFL + ΔFR = 12250 N
ΔFL = 12250 - 4375 = 7875 N
Answer:
Actually the same happens when the ray passes through optical centre. This can be observed in a thick lens. In thin lenses the perpendicular distance between extended incident ray and extended emergent ray is negligible. So we can say that light ray passes through optical centre without deviation.
Explanation:
The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.
<h3>Work done by the applied force</h3>
The area under force versus displacement graph is work done.
The total work done by pulling the object 7 m, can be grouped into two areas;
- First area, A1 = area of triangle from 0 m to 2.0 m
- Second area, A2 = area of trapezium, from 2.0 m to 7.0 m
A1 = ¹/₂ bh
A1 = ¹/₂ x (2) x (20)
A1 = 20 J
A2 = ¹/₂(large base + small base) x height
A2 = ¹/₂[(7 - 2) + (7-3)] x 50
A2 = ¹/₂(5 + 4) x 50
A2 = 225 J
<h3>Total work done </h3>
W = A1 + A2
W = 20 J + 225 J
W = 245 J
Learn more about work done here: brainly.com/question/8119756
1 - Skull
2 - Mandible
3 - Scapula
4 - Sternum
5 - Ulna
6 - Radius
7 - Pelvis
8 - Femur
9 - Patella
10 - Tibia
11 - Fibula
12 - Metatarsals
13 - Clavicle
14 - Ribs (rib cage)
15 - Humerus
16 - Spinal column
17 - Carpals
18 - Metacarpals
19 - Phalanges
20 - Tarsals
21 - Phalanges