Ask question

Ask question

All categories

2 years ago

13

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

on which you can either stand or walk. Suppose a speed ramp has a length of 121 m and is moving at a speed of 2.2 m/s relative to the ground. In addition, suppose you can cover this distance in 78 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 121 m using the speed ramp?

1 answer:

Volgvan2 years ago

8 0

Answer:

It takes you 32.27 seconds to travel 121 m using the speed ramp

Explanation:

<em>Lets explain how to solve the problem</em>

- The speed ramp has a length of 121 m and is moving at a speed of

2.2 m/s relative to the ground

- That means the speed of the ramp is 2.2 m/s

- You can cover the same distance in 78 seconds when walking on

the ground

<em>Lets find your speed on the ground</em>

Speed = Distance ÷ Time

The distance is 121 meters

The time is 78 seconds

Your speed on the ground = 121 ÷ 78 = 1.55 m/s

If you walk at the same rate with respect to the speed ramp that

you walk on the ground

That means you walk with speed 1.55 m/s and the ramp moves by

speed 2.2 m/s

So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s

Now we want to find the time you will take to travel 121 meters using

the speed ramp

Time = Distance ÷ speed

Distance = 121 meters

Speed 3.75 m/s

Time = 121 ÷ 3.75 = 32.27 seconds

It takes you 32.27 seconds to travel 121 m using the speed ramp

You might be interested in

Two particles carrying charges q1 and q2 are separated by a distance r and exert an electric force F⃗ E on each other. If q1 is

zepelin [54]

Answer:

q2 must also be doubled

r may also be halved

Explanation:

According to Coulumbs law

F= K q1 q2/r^2

If q1 is doubled, we must necessarily double q2 and r may also be halved in order to maintain F at the same value. Once the value of F is thus kept constant and E is also constant, the product FE must remain constant.

5 0

2 years ago

Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi

morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = $\dfrac{0.029158}{4.03176}$ = 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

4 0

2 years ago

Read 2 more answers

A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4

HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$2.8 \text{ m} = 0+\dfrac{gt^2}{2}$

$t = \sqrt{\dfrac{5.6}{g}}$

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g </em> (It is going up)

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$1.2\text{ m} = ut-\frac{1}{2}gt^2$

Substituting the value of <em>t</em> from the previous equation,

$1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}$

$u\sqrt{\dfrac{5.6}{g}} = 4.0$

Taking <em>g</em> = 9.8 m/s²,

$u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}$

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

$u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}$

The initial velocity is

$v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}$

4 0

2 years ago

To prevent collisions and violations at intersections that have traffic signals, use the _____ to ensure the intersection is cle

Ann [662]

The answer should be:
<span>To prevent collisions and violations at intersections that have traffic signals, use the d</span>elayed acceleration technique<span> to ensure the intersection is clear before you enter it.
Delayed acceleration technique refers to w</span><span>aiting to go through an intersection until you have a chance to scan for other vehicles.</span>

3 0

2 years ago

If one firecracker produces a sound intensity of 90db, what would be the intensity of a sound produced by 1000 firecrackers, all

Kryger [21]

Answer:

90db

Explanation:

The 1000, will produce sa.e intensity, since the firecrackers are made of same materials

5 0

2 years ago

Read 2 more answers