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Nadya [2.5K]
3 years ago
10

A moving walkway has a speed of 0.9 m/s to the east. A stationary observer sees a man walking on the walkway with a velocity of

1.1 m/s to the east. What is the man's velocity relative to the moving walkway?
Physics
1 answer:
Marina CMI [18]3 years ago
3 0
Since the man and the walkway has velocity (which is a vector) moving in the same direction, who would add the velocities together.

That means the relative velocity would be 2 m/s to the East

If you picture it in your mind, imagine walking down an escalator, you feel like you're moving faster than you really are.

In a different example, if the man was moving 0.9 m/s to the West, they are moving in opposite directions, which means you would subtract the 2 velocities.
You might be interested in
When we breathe, we inhale oxygen and exhale carbon dioxide plus water vapor. Which likely has more mass, the air that we inhale
Solnce55 [7]

Answer:

If the same volume of air is inhaled and exhaled, the air we breathe out normally weighs more than the air we breathe in.

Since the output from the body normally exceeds the input, breathing leads to weight loss.

Explanation:

If equal volumes of gas is inhaled and exhaled, the exhaled gas is heavier.

The inhaled gas contains Oxygen and majorly Nitrogen.

The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.

So, basically, the body exchanges O₂ with CO₂ and H₂O (and some other unwanted gases in the body) in a composition that CO₂, the heavier gas of the ones mentioned here, is prominent.

So, because the mass leaving the body is more than the mass entering, breathing leads to a loss of weight. This is one of the reasons why we need food for sustenance. Breathing alone will wear one out.

5 0
3 years ago
Consider a 24.0 V storage battery that can transfer, over the course of its useful lifetime, a total charge equivalent to 340000
xeze [42]

Answer:

Explanation:

Given that,

Potential difference of battery is

V = 24V

Total charge battery can transferred

q = 340,000C

Work done by the battery?

Work done is given as

W = qV

Where q is charge in Columbs

V is potential difference in Volts

Then, W = qV

W = 340,000×24

W = 8,160,000 J

Work done by the battery over it's useful life time is 8,160,000J

5 0
3 years ago
An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

  • a) \vec{v} = 70·i
  • b) The package will reach ground in approximately <u>12.77 seconds</u>.
  • c) The speed of the package as it lands is approximately <u>145.51 m/s</u>.
  • d) The path of the package based on a stationary frame of reference is <u>parabolic</u>
  • e) The path of the package as seen from the plane is <u>directly vertical</u> downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, \vec{v} = 70·i

b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;

\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}

Which gives;

\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}

The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>

c) The vertical velocity just before the package reaches the ground, v_y, is given as follows;

v_y^2 = 2·g·h

Therefore;

v_y = √(2·g·h)

Which gives;

v_y = √(2 × 9.81 × 800) ≈ 125.28

v_y ≈ 125.28 m/s

Which gives; \vec{v} = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ <u>143.51 m/s</u>

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating <u>directly vertical</u>

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0
2 years ago
If you lift a five pound object 18 inches how many joules of energy did yo use?
kogti [31]
The work is 90 as 5 times 18

4 0
3 years ago
For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red
barxatty [35]

Answer:

Explanation:

To calculate the red shift you use the following formula:

z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

z=\sqrt{\frac{c+v}{c-v}}-1

- A red dwarf moving away from Earth at 39.1 km/s :

z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}

- A blue dwarf moving away from Earth at 25.9 km/sz=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}

- A red dwarf moving transversely at 14.1 km/s

z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}

6 0
4 years ago
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