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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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zhuklara [117]

Answer:

Dance studio

Explanation:

Martial art use to defend ourself from any dangerous. Dance is a way to learn it

3 0
2 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
Explain why the shortening velocity became slower as the load became heavier in this experiment. how well did the results compar
valkas [14]
<span>The shortening velocity refers to the speed of the contraction from the muscle shortening while lifting a load. Maximal shortening velocity is only attained with a minimal load. With a light load, the shortening velocity is at its Maximal shortening velocity. When the weight is heavy, the speed in which the muscle lifts the weight decreases in speed at a slower velocity.</span>
6 0
3 years ago
A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc
julsineya [31]

Answer:

15

Explanation:

mass, M = 5Kg

horizontal force, F_h = 40N

acceleration, a =5 m/s^2

frictional force, F_f =?

net force = ma

net force = F_h - F_f = 40N - F_f

40  - F_f = 5 x 5

- F_f = 25 - 40

multiply both side by -1

F_f = 40 - 25 = 15

the frictional force is 15N

4 0
3 years ago
Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

\lambda =2(5.00)=10.0 m

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]

And by comparing it with the general equation of a wave:

y(t)=A sin (\frac{2\pi}{T} t)

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

6 0
3 years ago
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