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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
Physics
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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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0
3 years ago
Please answer me fast ​
meriva

Answer: i think c

Explanation:QA: “What is ordinary glass made of ?”

Glass is mostly silica, or silicon dioxide, present as quartz in many types of sand. Pure silica forms a highly transparent glass, but has a very high melting or softening temperature, around 1700°C. Even at such high temperatures it is highly viscous and difficult to work. Its use is largely confined to applications requiring high transparency to ultra-violet and infra-red radiation, stability at elevated temperatures or low thermal expansion coefficient.

“Ordinary glass” windows and drinking vessels are typically made from soda-lime glass, containing silica with around 25% sodium, calcium and other oxides, which together reduce the softening temperature to roughly 500–600°C

6 0
3 years ago
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
NARA [144]

Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = 6\times 10^6\ J

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s

The time taken to lift the load is 141.26626 seconds

5 0
3 years ago
A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =
Liula [17]
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam. 
1)What is the force the left support exerts on the beam? 
2)What is the force the right support exerts on the beam? 
3)How much extra mass could the gymnast hold before the beam begins to tip? 
Now the gymnast (not holding any additional mass) walks directly above the right support. 

4)What is the force the left support exerts on the beam? 
5)What is the force the right support exerts on the beam?</span>
6 0
3 years ago
The open ocean, sea floor, and coral reefs are all examples of habitats with high rates of primary production.
lisov135 [29]

the answer you are seeking is false

 



7 0
3 years ago
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