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Step2247 [10]
3 years ago
5

What is the approximate weight of the air inside the tire in English Engineering Units (tire outside diameter = 49", rim diamete

r-22", tire width-19" see: tire side view. ndf?
Physics
1 answer:
Mademuasel [1]3 years ago
3 0

Answer:

1.265 Pounds

Explanation:

Data provided:

Tire outside diameter = 49"

Rim diameter = 22"

Tire width = 19"

Now,

1" = 0.0254 m

thus,

Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 × 0.0254 = 0.6223 m

Rim radius, r₂   = 22" / 2 = 11" = 0.2794 m

Tire width, d = 19" = 0.4826 m

Now,

Volume of the tire = π ( r₁² - r₂² ) × d

on substituting the values, we get

Volume of air in the tire = π (  0.6223² - 0.2794² ) × 0.4826 = 0.46877 m³

Also,

Density of air = 1.225 kg/m³

thus,

weight of the air in the tire = Density of air × Volume air in the tire

or

weight of the air in the tire = 1.225 × 0.46877 = 0.5742 kg

also,

1 kg = 2.204 pounds

Hence,

0.5742 kg = 0.5742 × 2.204  = 1.265 Pounds

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Kryger [21]

Answer:

yes kinetic energy will be affected it will get increased 16 times.

Explanation:

4 0
3 years ago
If you used 650 W of power lifting a 300 N weight in 2 seconds how high did you lift?
dimulka [17.4K]

Answer:

4.33m

Explanation:

Power = work done/ time

work done = power × time =650 × 2 = 1300J

work done = force × distance

distance = work done/force

distance = 1300/300 = 4.33m

7 0
3 years ago
What is the scientific definition of Horizontal component and vertical component.
Talja [164]
I think it’s true but I’m not sure
3 0
4 years ago
Sherlock Holmes examines a clue by holding his magnifying glass (with
Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

M=-\frac{q}{p}=-\frac{-30}{10}=3

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0
4 years ago
b) The depth of a wrecked ship is 1000 m below sea level. A ship transmits a wave of frequency 50 kHz and receives an echo 2.4 s
alexandr402 [8]

The speed of the wave in water is 833.33 m/s

The given parameters;

  • depth of the wrecked ship, d = 1000 m
  • frequency of the wave transmitted, f = 50 kHz
  • time of motion of the echo, t = 2.4 s

The speed of the wave in water is calculated  by applying echo equation as shown below;

2d = velocity \times time\\\\velocity = \frac{2d}{time} \\\\velocity =  \frac{2\times 1000 \ m}{2.4} \\\\velocity = 833.33 \ m/s

Thus, the speed of the wave in water is 833.33 m/s

Learn more here: brainly.com/question/23433611

4 0
3 years ago
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