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BigorU [14]
2 years ago
12

You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly

breaks?
Physics
2 answers:
Viefleur [7K]2 years ago
7 0

A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

Alenkinab [10]2 years ago
6 0

The bathroom scale gives the reading as zero as soon as the cable of the elevator breaks because the elevator starts to fall with the acceleration same as the acceleration due to gravity.

Further Explanation:

The bathroom scale kept in an elevator shows the weight of the person standing on the scale. The weight of a person is the measure of the force experienced by the person due to the acceleration due to gravity of Earth.

Since the scale is kept in an elevator, the reading of the scale is affected by the acceleration of the elevator as it exerts a pseudo force on the person standing on the scale in opposite direction.

If the cables holding the elevator are broken suddenly, the elevators starts to fall will the acceleration due to gravity and the person on the scale experiences weightlessness inside the elevator as the pseudo force acting in the opposite direction is equal to his weight. Therefore, the bathroom scale will give the reading zero as the cable of the elevator breaks.

Learn More:

  1. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 n brainly.com/question/7031524
  2. What type of mirror do dentist use brainly.com/question/997618
  3. Which of the following is not a component of a lever brainly.com/question/1073452

Answer Details:

Grade: High School

Chapter: Force and acceleration

Subject: Physics

Keywords:  Bathroom scale, moving elevator, scale reading, acceleration due to gravity, suddenly breaks, weightlessness, pseudo force, opposite direction.

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Amanda [17]

Answer:

amount of energy  = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

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8 0
2 years ago
Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1
7nadin3 [17]

Answer:

A). σ = 3.823 x 10^{-5} C^{2}/N-m^{2}

B). \sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C). U=10.322 J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

\sigma =k.\varepsilon _{0}.E

where, E is resultant electric field = 1.2 x 10^{6} V/m

           \varepsilon _{0} is permittivity of free space = 8.85 x 10^{-12} C^{2}/N-m^{2}

           k is dielectric constant = 3.6

∴\sigma =k.\varepsilon _{0}.E

                     = 3.6 x 8.85 x10^{-12} x 1.2 x 10^{6}

                    = 3.823 x 10^{-5} C^{2}/N-m^{2}

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )

\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )

\sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C).

Area of the plate, A = 2.5 cm^{2}

                                 = 2.5 x 10^{-4}m^{2}

diameter of the plate, d = 1.8 mm

                                        = 1800 m

∴ Total energy stored in the capacitor

U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad

U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800

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4 0
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A planar loop consisting of four turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field
spayn [35]

Answer:

0.6A

Explanation:

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From Faraday' s law of induction EMF(E)= change in magnetic field/time

    E= 15mT/5ms

Note, that one weber per second is equivalent to one volt.

= 3V

from Ohm's law I =E/R  

=3/5 =0.6A

 

7 0
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A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0
1 year ago
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Alenkinab [10]

Answer:

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Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

6 0
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