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Zepler [3.9K]
3 years ago
5

Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh

t is 690 N (155 lbs) and the elevator moves upward at 0.25 g and then down at 0.25 g, what is the difference between the up and down scale readings?
Physics
1 answer:
GrogVix [38]3 years ago
4 0

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=N_1-mg=N_1-690

Now, from Newton's second law:

F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=mg-N_2=690-N_2

Now, from Newton's second law:

F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

Difference=N_1-N_2=862.5-517.5=345\ N

Therefore, the difference between the up and down scale readings is 345 N.

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Veseljchak [2.6K]

Answer:

a) 0, = -0.33 us

b) 140m

c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.

Explanation:

a)

the lorentz factor expression is written as;

y = 1₀ / √(1 - (v²/c²))

where v  is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

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y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

1 - the coordinates  of the first event, the s' frame of reference is,

x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

= 0

2 - the coordinates of the second event, the s ' frame of reference is'

x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

using speed of light c as 3*10^8

1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

= -0.33 us

b)

distance between

delltaX' = X2' - X1'

= 140m - 0

= 140m

c)

No, The event are not simultaneous i.e they did not occur at the same time.

the second even (-0.33 us) occurs 0.33 us earlier than the first event.

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3 years ago
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Reil [10]

Answer:

\Delta l=0.015m

Explanation:

We have given initial length of the steel guitar l = 1 m

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Young's modulus \gamma=2\times 10^{11}Pa

Force F = 1500 N

So stress =\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa

We know that young's modulus =\frac{stress}{strain}

So 2\times 10^{11}=\frac{3\times 10^{9}}{strain}

strain=1.5\times 10^{-2}=0.015m

Now strain =\frac{\Delta l}{l}

0.015=\frac{\Delta l}{1}

\Delta l=0.015m

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mamaluj [8]

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Unknown:

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The force exerted by all the four feet is 0.16m²

the area of each feet = \frac{0.16}{4} = 0.04m²

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</span>
<span>for f=4200 Hz </span>
<span>λ= v/f=337/4200= 0.08 m </span>
<span>So max. wavelength is 12.036 m and </span>
<span>Min Wavelength is 0.08 m </span>
<span>So the range is between .08 m and 12.036 m
</span>Hope this helps. 
4 0
3 years ago
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