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Margaret [11]
2 years ago
3

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 130 lb as the door closes. The reading varies between 100 lb and 160 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:
Katena32 [7]2 years ago
5 0

Answer:

Both have the magnitude of 0.23g, or 2.264 m/s2

Explanation:

So the scale would read 160lb when the elevator accelerates upward and 100lb when the elevator accelerate downward

As 160 / 130 = 1.23, you are experiencing 1.23 g when accelerate upwards. The magnitude of the acceleration upward would be 1.23g - g = 0.23 g, or 9.81*0.23 = 2.264 m/s2.

Similarly, 100 / 130 = 0.77, you are experiencing 0.77 g when accelerate downwards. The magnitude of the acceleration downward would be g - 0.77g =  0.23 g, or 9.81*0.23 = 2.264 m/s2.

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You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant spee
Lina20 [59]

Answer:

40.13491 m/s

Explanation:

v_r =  My speed = 35 m/s

v = Speed of sound in air = 343 Hz

v_s = Speed of the police car

When the car is approaching

f=f'\dfrac{v-v_r}{v-v_s}\\\Rightarrow 1340=f'\dfrac{343-35}{343-v_s}

When the car is receding

f=f'\dfrac{v+v_r}{v+v_s}\\\Rightarrow 1300=f'\dfrac{343+35}{343+v_s}

Dividing the equations

\dfrac{1340}{1300}=\dfrac{f'\dfrac{343-35}{343-v_s}}{f'\dfrac{343+35}{343+v_s}}\\\Rightarrow \dfrac{1340}{1300}=\dfrac{22\left(v_s+343\right)}{27\left(-v_s+343\right)}\\\Rightarrow -36180v_s+12409740-12409740=28600v_s+9809800-12409740\\\Rightarrow \frac{-64780v_s}{-64780}=\frac{-2599940}{-64780}\\\Rightarrow v_s=\frac{129997}{3239}\\\Rightarrow v_s=40.13491\ m/s

The speed of the police car is 40.13491 m/s

5 0
2 years ago
A ball of mass 0.4 kg is initially at rest on the ground. It is kicked and leaves the kicker's foot with a speed of 5.0 m/s in a
yawa3891 [41]

Answer:

the answer the correct  is 3

Explanation:

Let's use the relationship between momentum and momentum

         I = Δp

         I = m v_{f} - m v₀

     

Let's calculate

         I = 0.4 5.0 - 0

         I = 2.0 N s

By Newton's law of action and reaction the force on the ball is equal to the force that the ball exerts on the foot, therefore the impulse on the foot of equal magnitude, but in the opposite direction

        I = 2.0 Ns with 60°

When reviewing the answer the correct  is 3

4 0
2 years ago
A brick is resting on a rough incline as shown in the figure. The friction force acting on the brick, along the incline, is
Tasya [4]
B) equal to the gravitational force of the brick
6 0
2 years ago
Read 2 more answers
The length of the side of a cube having a density of 12.6 g/ml and a mass of 7.65 g is __________ cm.
qaws [65]

Density is the characteristic property of a substance.  It is the measure of mass of the substance divided by its volume (density= mass/volume). Manipulate the given formula to come up with the formula for the volume. Therefore, volume is equals to mass of a substance divided by its density (Vol= mass/density). Given 12.6 g/ml as density and 7.65 g mass, volume is equals to 0.60714 ml, since 1 ml = 1cm^3, volume is equals to 0.60714 cm^3 then extract the cube root of the volume to get the length of the cube in cm which is equal to 0.84677 cm.




4 0
2 years ago
On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s
xenn [34]
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of T=31^{\circ}C:
v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
v= \frac{2L}{t}
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}=  804.1 m
6 0
2 years ago
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