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suter [353]
1 year ago
10

A 1500 kg elevator, suspended by a single cable with tension 16.0 kN, is measured to be moving upward at 1.2 m/s. Air resistance

is negligible.1. We want to find the elevator’s speed after traveling upward 10.0 m. This could be solved with Newton's laws and kinematics, but we will use the energy principle. How much work does tension do on the elevator as it rises 10.0 m?2. How much work does gravity do on the elevator as it rises 10.0 m?3. What is the elevator's kinetic energy after rising 10.0 m?4. What is the elevator's speed after rising 10.0 m?
Physics
1 answer:
kirill115 [55]1 year ago
6 0

Tension in the Cable is 0.87 m/s6^2, Elevator's speed after it has moved 10m is 1.6*10^5 N, Work done by gravity is  1.47 * 10^3 N, Elevator Kinetic Energy is 13897.5J and Elevator Speed after rising to 10m is 4.312 m/s.

Tension is a pulling force that operates in one dimension along the cables' axes in the opposite direction from the direction of the applied force. The combined weight of the elevator box and the passenger riding inside it, in the case of an elevator, provides the pulling force in the cables is called Tension.

A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. The type of motion can be vibration, translation, rotation around an axis, or any combination of these. Kinetic energy is a type of energy that an item or particle possesses as a result of motion.

We know that,

Tension in the Cable

T = m(g+a)  = g+a = T/m = 16 * 103 / 1500 = 10.67 m/s2

a = 10.67 - 9.8 = 0.87 m/s6^2

Elevator's speed after it has moved 10m.

U^2 = u^2 +2as

= 1.22 +2*0.87*10

=1.6*10^5 N

Work done by gravity  = mg * 10 = 14700 * 10 = 1.47 * 10^3 N

Elevator Kinetic Energy = 1/2 mv^2 = 1/2*1500*18.53 = 13897.5J

Elevator Speed after rising to 10m ,

U^2 = u^2 +2as =  1.2 +2*0.87*10 = 18.6

U =(18.6)^1/2=4.312 m/s

Learn more about Kinetic Energy here

brainly.com/question/26472013

#SPJ4

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Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
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Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

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V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

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equating both expressions, we have

V₁/R₁ = V₂/R₂

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dividing through by 2.0 V, we have

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= 3.0 Ω

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= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

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