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Sliva [168]
2 years ago
4

A simple elevator ride can teach you quite a bit about the normal force as this rider below can (hopefully) tell you. There are

three different scenarios given, detailing the rider\'s experience in an unnamed hotel. For each scenario, calculate the normal force, FN,1-3, acting on the rider if his mass is m = 76.6 kg and the acceleration due to gravity g = 9.81 m/s2. In scenario 1, the elevator has constant velocity. In scenario 2 the elevator is moving with upward acceleration a2 = 4.84 m/s2. Finally, in scenario 3, unfortunately for the rider, the cable breaks and the elevator accelerates downward at a3 = 9.81 m/s2.
FN1= _______
FN2=_______
FN3=___________
Physics
1 answer:
Alina [70]2 years ago
6 0

Answer:

FN1 = 751.5 N  

FN2 = 1122.2 N  

FN3 = 0

Explanation:

Scenario 1 :

• The elevator has constant velocity.

  • The normal force, can adopt any value, as needed by Newton's 2nd Law, in order to fit this general expression:
  • Fnet = m*a
  • In the first scenario, as the elevator is moving at a constant speed, this means that no external net force is present.
  • The two forces that act on the rider, are gravity (always present, downward) and the normal force, as follows:
  • Fnet = Fn - m*g = m*a  (taking the upward direction as positive)

For scenario 1:

Fnet = 0 ⇒ Fn = m*g = 76.6 kg * 9.81 m/s² = 751. 5 N

Scenario 2:

  • In this scenario, the elevator has an upward acceleration of 4.84 m/s², so the Newton's 2nd Law is as follows:
  • Fnet = FN - m*g = m*a  
  • ⇒ FN = m* ( g+ a) = 76.6 kg* (9.81 m/s² + 4.84 m/s²) = 1,122.2 N

Scenario 3 :

  • As the elevator is in free fall, this means that a = -g, so, in this condition, the normal force is just zero, as it can be seen from the following equation:
  • FN-mg = m*a  
  • If a = -g,  
  • ⇒ FN -mg = -mg ⇒ FN=0

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gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

And when length is L_2 time period T_2=1sec

We know that time period is given by

T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

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B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find
lozanna [386]

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

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Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

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Which two pieces of data indicate that Uranus resides in the outer region of the solar system
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Answer:

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14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?
Vladimir79 [104]
The relationship between resistance R and resistivity \rho is
R= \frac{\rho L}{A}
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m
and the cross section is
A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2

From the first equation, we can then find the length of the wire when R=4.8 \Omega (copper resistivity: \rho = 1.724 \cdot 10^{-8} \Omega m)
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Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual
kotegsom [21]

Answer:

The amount of heat required is H_t =  1.37 *10^{6} \ J

Explanation:

From the question we are told that

The mass of water is m_w  =  20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g

The temperature of the water before drinking is T_w  =  3.8 ^oC

The temperature of the body is T_b  =  36.6^oC

Generally the amount of heat required to move the water from its former temperature to the body temperature is

H=  m_w  *  c_w * \Delta T

Here c_w is the specific heat of water with value c_w = 4.18 J/g^oC

So

H=   5.7 *10^2 * 4.18 * (36.6 - 3.8)

=> H= 7.8 *10^{4} \  J

Generally the no of mole of sweat present mass of water is

n = \frac{m_w}{Z_s}

Here Z_w is the molar mass of sweat with value

Z_w =  18.015 g/mol

=> n = \frac{5.7 *10^2}{18.015}

=> n = 31.6 \  moles

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

H_v  =  n  *  L_v

Here L_v is the latent heat of vaporization with value L_v  = 7 *10^{3} J/mol

=> H_v  =  31.6 * 7 *10^{3}

=> H_v  = 1.29 *10^{6} \  J

Generally the overall amount of heat energy required is

H_t =  H +  H_v

=> H_t =  7.8 *10^{4} +  1.29 *10^{6}

=> H_t =  1.37 *10^{6} \ J

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2 years ago
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