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Zinaida
1 month ago
11

A moving sidewalk 95 m in length carries passengers at a speed of 0.53 m/s. One passenger has a normal walking speed of 1.24 m/s

. (a) If the passenger stands on the sidewalk without walking, how long does it take her to travel the length of the sidewalk? (b) If she walks at her normal walking speed on the sidewalk, how long does it take to travel the full length? (c) When she reaches the end of the sidewalk, she suddenly realizes that she left a package at the opposite end. She walks rapidly back along the sidewalk at double her normal walking speed to retrieve the package. How long does it take her to reach the package?
Physics
1 answer:
Maru [3.3K]1 month ago
3 0
Answer: a) t = 1.8 x 10^2 seconds; b) t = 54 seconds; c) t = 49 seconds. Explanation: a) To determine the time of a stationary passenger on the sidewalk, we use the position formula. Given the constant speed of the walkway, we can calculate the time taken for set distances accordingly. This calculation extends into cases where combined velocities for walking are involved in subsequent queries.
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"if a stream flow measures 12 meters in 60 seconds, what is the stream's average rate of flow?"
Yuliya22 [3333]
Discharge refers to the volume of water flowing down a river or stream within a specific timeframe, typically measured in cubic feet per second or gallons per day. Generally, the discharge of a river is calculated by taking the product of the cross-sectional area of water in the channel and the average velocity of water at that section: discharge = area * velocity. In this instance, the result is 0.2 m/s.
7 0
1 month ago
A rope connects boat A to boat B. Boat A starts from rest and accelerates to a speed of 9.5 m/s in a time t = 47 s. The mass of
ValentinkaMS [3465]

Answer: 339.148N

Explanation:

Given data:

Time (t) = 47s

Initial speed (U) = 0m/s

Final speed (V) = 9.5m/s

Mass of B = 540kg

Frictional force on B = 230N

Since both boats are linked, movement of A causes B to move as well.

What is the acceleration of boat A?

Applying the motion formula:

V = u + at

9.5 = 0 + a * 47

a = 9.5 / 47

a = 0.2021 m/s²

To determine the force necessary to accelerate boat B, as both boats experience the same force:

F = Mass * acceleration

F = 540 * 0.2021 = 109.14N

Given that there is a frictional force of 230N acting on boat B, the overall force (Tension) becomes:

Tension = frictional force + applied force = (109.14 + 230)N = 339.148N

7 0
2 months ago
the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point
serg [3582]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We are tasked with finding the pressure here

Applying the pressure formula

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v refers to velocity

Insert the values into the equation

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

Therefore, the pressure at this moment is 0.875 mPa

5 0
3 months ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
Keith_Richards [3271]

Answer:

The flux across the cube's surface is 2.314\ Nm^{2}/C.

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = 8.0\times 10^{- 2}\ m.

Volume charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}.

Now,

To find the electric flux:

\phi = \frac{q}{\epsilon_{o}}

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of vacuum.

The volume charge density for this scenario is described by:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}

Cube volume, V = a^{3}.

Thus,

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}.

The total charge can be derived from equation (2):

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}.

q = 2.048\times 10^{-11}\ F = 20.48\ pF.

Now, insert the value of 'q' into equation (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C.

5 0
2 months ago
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