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PilotLPTM [1.2K]
3 years ago
13

A bridge is supported by two piers located 14 meters apart. Both the left and right piers provide an upward force on the bridge,

labeled FL and FR respectively. (a) If a 1250 kg car comes to rest at a point 5 meters from the left pier, how much force (in N) will the bridge provide to the left and right piers? (Enter the magnitudes.)
Physics
1 answer:
almond37 [142]3 years ago
5 0

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

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sveta [45]

The work done on the car is -20 J.

Work done on the car is negative, meaning that the car actually does work on the external system.

<h3>Energy and law of conservation of energy</h3>
  • Energy is the ability to do work
  • the law of conservation of energy states that the total energy in a system is conserved

From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.

  • Initial energy = 100 J
  • Initial energy = Final energy - work done on car
  • Final Energy = Work done on car + initial energy

80J = Work done on car + 100 J

Work done on car = 80 - 100J

Work done on car = -20 J

Hence, the work done on the car is -20 J

Work done on car is negative.

Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.

Learn more about energy and work at: brainly.com/question/13387946

8 0
2 years ago
Select the correct answer.
lutik1710 [3]
I’m sorry i haven’t found the answer to this
8 0
3 years ago
Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th
katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

F=2*10^{4}N

Explanation:

From the question we are told that:

Mass m=0.500kg

Initial Velocity V=15.0m/s

Distance x=2.80mm=>0.00280m

Diameter d=2.50mm=>0.00250m

Length l=6.00cm=>0.6m

Generally the equation for Force is mathematically given by

 F=\frac{mv^2}{2d}

 F=\frac{0.500*15^2}{2.80*10^{-3}}

 F=2*10^{4}N

6 0
3 years ago
A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. The trains collide. After the collisi
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Explanation:

Momentum = mass × speed

p = (30 kg) (3 m/s)

p = 90 kg m/s

7 0
3 years ago
1. The gravitational force acting on a falling body and its weight is constant. But the law of universal gravitation tells us th
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It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.
3 0
3 years ago
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