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Brilliant_brown [7]

3 years ago

13

When the moving sidewalk at the airport is broken, as it often seems to be, it takes you 44 s to walk from your gate to baggage

claim. When it is working and you stand on the moving sidewalk the entire way, without walking, it takes 90 s to travel the same distance

1 answer:

Law Incorporation [45]3 years ago

3 0

Answer:

a) 29.55 sec

Explanation:

Let the total distance be s feet.

Speed while walking = s/44 ft/sec.

Speed on sidewalk = s/90 ft/sec.

Total speed while walking on moving sidewalk

= s/44+s/90 = (90s+44s)/44×90

=(134x)/3960 ft/sec

= x/29.55 ft per sec.

Hence your travel time will be 29.55 secs.

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3 years ago

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

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So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

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3 0

3 years ago

grandymaker [24]

The answer is B tell me if I am wrong.

6 0

3 years ago

3 An un calibrated mercury in glass thermometer immersed in melting ice. The length of the mercury thread is 25 mm when the ther

sammy [17]

Answer:

25 mm = 0 deg C

200 mm = 100 deg C

200 - 25 = 175 = change in thread per 100 deg C

95 - 25 = 70 mm - change in thread from 0 deg C

70 / 175 * 100 = 40 deg C final temperature at 95 mm

5 0

3 years ago

2. A portion of the body receives 0.15 mGy from radiation with a quality factor Q = 6 and 0.22 mGy from radiation with Q = 10. (

DiKsa [7]

Answer with Explanation:

We are given that

$D_1=0.15mGy$

$D_2=0.22mGy$

$Q_1=6$

$Q_2=10$

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Total dose= $D=D_1+D_2$

Using the formula then, we get

$D=0.15+0.22$

$D=0.37mGy$

b.We have to find the total dose equivalent

Total dose equivalent=H= $D_1Q_1+D_2Q_2$

Using the formula

$H=0.15(6)+0.22(10)$

H=3.1mSv

7 0

3 years ago