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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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The time it takes a car to attain a speed of 30 m/s when accelerating from rest at 2 m/s2 is?
Blizzard [7]

Explanation:

u = 0m/s

v = 30m/s

a = 2m/s²

Using Kinematics, we have v = u + at.

Therefore t = (v - u) / a

= (30 - 0) / (2) = 15 seconds. (B)

7 0
3 years ago
With time running out in a game, Rachel runs towards the basket at a speed of 2.5 meters per second and from half-court, launche
finlep [7]

Answer:

Rachel(2.5,0)

ball(6.5,4.7)

b.R=10.15m/s, 27.57deg

Explanation:

The reference angle of Rachel is 00^{0} while the ball is at 36^{0}

resolving rachel's speed to the horizontal, we have

Ux=2.5cos0

Ux=2.5m/s

resolving rachel's speed to the vertical we have,

Uy=2.5sin0

Uy=0

for the ball

resolving the speed to its horizontal component

Ux=8cos36

Ux=6.5m/s

Uy=8sin36

Uy=4.7m/s

Rachel(2.5,0)

ball(6.5,4.7)

To get the resultant of their speed

Add the horizontal speed of rachel to that of the ball to get the total horizontal speed

Add the vertical speed of rachel and the ball to get the total vertical speed component

TUx=2.5+6.5=9M/S

TUy=0+4.7=4.7m/s

R=\sqrt{(TUx^2+TUy^2}

R=\sqrt{(9^2+4.7^2}

R=\sqrt{(103)}

R=10.15m/s

the direction

tan\alpha=TUy/TUx

tan\alpha=4.7/9

\alpha=tan^-1(0.522)

\alpha=27.57deg

4 0
3 years ago
Two nuclear reactors provide 3200 MW of power.If the transmission system loses 5.1% of the energy produced,how much power from t
Annette [7]

The power that the customers receive is 163.2 MW

Explanation:

The efficiency of a system is given by

\eta = \frac{P_{out}}{P_{in}}

where

P_{out} is the power in output

P_{in} is the power in input

For the nuclear reactors in this problem, we have:

P_{in} = 3200 MW (power generated by the reactors)

\eta=0.051 (efficiency is 5.1%)

Therefore, the power that reaches the customers is:

P_{out} = \eta P_{in} = (0.051)(3200)=163.2 MW

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

6 0
3 years ago
Two small, positively charged spheres have a combined charge of 5.23 x 10^-5 C. If each sphere is repelled from the other by an
pshichka [43]

Answer:

The smallest charge of the dial is 2.46 10-5 C

Explanation:

The Coulomb force is responsible for the electroactive repulsion, the equation that describes it is

       

          F = k q1 q2 / r²

Where K is the Coulomb constant that is worth 8.99109 N m² / C², q is the electric charge of each sphere and r is the distance between them.

They also give us the condition that the sum of the charge is 5.23 10-5 C

          Qt = q1 + q2 = 5.23 10⁻⁵ C

Let's replace in the Coulomb equation, let's clear and calculate

       

         F = k (Qt -q2) q2 / r²

         F = k q22 / r² - k Qt q2 / r²

         1.0 = 8.99 10⁹ q2² /2.04² - 8.99 10⁹ 5.23 10⁻⁵ q2 / 2.04²

         1.0 = 2.16 10⁹ q2²2 - 11.30 10⁴ q2

         0 = 2.16 109 q2² - 11.30 10⁴ q2 -1.0                (* 1/2.16 109)

          0 = q2² - 1.05 10⁻⁵ q2 - 0.463 10⁻⁹

Let's solve the second degree equation for q2

         q2 = 1.05 10⁻⁵ ±√[(1.05 10⁻⁵)² - 4 1 (-0.463 10⁻⁹)] / 2

         q2 = 1.05 10⁻⁵ ±√ [1.10 10⁻¹⁰ + 18.52 10⁻¹⁰] / 2

         q2 = {1.05 10⁻⁵ ± 4.43 10⁻⁵} / 2

The solutions are

        q2 ’= 2.74 10-5 C

        q2 ’’ = -1.69 10-5 C

As the problem tells us that the spheres are positively charged, the correct solution is 2.74 10-5 C, let's see the charge of the other sphere

                  Qt = q1 + q2 ’

                  q1 = Qt -q2 ’

                  q1 = 5.23 10-5 - 2.74 10-5

                  q1 = 2.46 10-5 C

The smallest charge of the dial is 2.46 10-5 C

5 0
4 years ago
In an electron cloud, an electron farther east away from the nucleus has?
vladimir2022 [97]

An electron that is far away from the nucleus have higher energy than an electron near the nucleus. Nucleus are positively charged and those electrons near it get attracted; those electrons gain kinetic energy hence reducing their internal energy. The electrons far from nucleus have low kinetic energy hence more internal energy.

8 0
3 years ago
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