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jolli1 [7]
3 years ago
13

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit

h the ground. Determine the weight of the automobile.
Physics
1 answer:
Stels [109]3 years ago
7 0

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, P=2.2\times 10^5\ Pa

Area of each tire, A=0.023\ m^2

Area of 4 tires,  A=0.092\ m^2

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

P=\dfrac{F}{A}

F=P\times A

F=2.2\times 10^5\times 0.092

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

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As per above given data

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now in order to find the change in velocity

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\Delta v = -38.1 km/s

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Part b)

Now we need to find acceleration

acceleration is given by formula

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given that

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a = - 5.84 * 10^{-4}m/s^2

so above is the acceleration

4 0
3 years ago
Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
Tanzania [10]
If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.

How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.

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Answer:

Monochronic.

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3 years ago
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
6 0
2 years ago
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