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valkas [14]
2 years ago
7

You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visit

ing New York, you take your bathroom scale into the elevator and stand on them. The scales read 140 lb as the door closes. The reading varies between 120 lb and 170 lb as the elevator travels 101 floors.i)What is the maximum acceleration upward?ii)What is the maximum magnitude of the acceleration downward?
Physics
1 answer:
love history [14]2 years ago
7 0

To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N

120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N

170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

F = 756 - 622.3

F = 133.7N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{133.7}{63.5}

a = 2.1m/s^2

PART B) For the maximum magnitude of the acceleration downward we have that:

F = 622.3 - 533

F = 89.3N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{89.3}{63.5}

a = 2.1m/s^2

a = 1.04 m/s^2

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Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. D
Ymorist [56]

Answer : The rate of heat transfer to the water is, 37.92 kJ/min

Explanation : Given,

Time = 10 min

Mass of water = 200 g

Latent heat of fusion of water = 334 J/g

Latent heat of vaporization of water = 2230 J/g

Now we have to calculate the rate of heat transfer to the water.

Q=\frac{m\times (L_v-L_f)}{t}

Now put all the given values in the above formula, we get:

Q=\frac{200g\times (2230-334)J/g}{10min}

Q=37920J/min=37.92kJ/min

Thus, the rate of heat transfer to the water is, 37.92 kJ/min

4 0
2 years ago
An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with
Anna71 [15]

(D) The gravitational force between the astronaut and the asteroid.

Reason :

All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.

6 0
2 years ago
Question 2 (1 point)
love history [14]

Answer:

neutral

Explanation:

3p - 3e = 0 and that leaves 2 neutrons so it will be neutral

6 0
2 years ago
Read 2 more answers
A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with
Lana71 [14]

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        Em_{f} = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     Em_{f} = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)

7 0
2 years ago
What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade?
kaheart [24]

Incomplete question.The complete question is here

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.

Answer:

F= 0.034 N

Explanation:

Given Data

Outer=2 cm

Edge of blade=21 cm

Mass=7.7 g

Length of blade=21 cm

The last 2cm is treated as if they were at a point 20cm from the center of rotation

To Find

Force=?

Solution

Convert the given frequency to angular frequency

ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)

ω= 3/2*π rad/sec

Now to find centripetal force.

F = m×v²/r

F= m×ω²×r

Put the data

F = 0.0077 kg × (3/2×π rad/sec )²× 0.20 m

F= 0.034 N

3 0
2 years ago
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