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vesna_86 [32]
3 years ago
6

An automobile tire having a temperature of −1.5 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 26 lb/in2 . What

would be the gauge pressure in the tire when its temperature rises to 40◦C? For simplicity, assume that the volume of the tire remains constant, that the air does not leak out and that the atmospheric pressure remains constant at 14.7 lb/in2 . Answer in units of lb/in2 .
Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:32.22\ lb/in^2

Explanation:

Given

Temperature on the hot day T_i=-1.5^{\circ} C\approx 271.5\ K

Gauge Pressure P_i=26\ lb/in^2

When Temperature rises to T_f=40^{\circ}C\approx 313\ K on a hot day then

absolute pressure  is let say P_f

Using ideal gas equation we can write

\Rightarrow \frac{PV}{T}=constant

as volume is constant therefore

\Rightarrow \frac{P}{T}=constant

\Rightarrow \frac{26+14.7}{271.5}=\frac{P_f}{313}

\Rightarrow P_f=46.92\ lb/in^2

Gauge Pressure is given by

\Rightarrow 46.92-14.7=32.22\ lb/in^2

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You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler
lbvjy [14]
In a stationary situation, the weight of person is
W=mg=(62 kg)(9.81 m/s^2) = 608.2 N
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
\sum F = ma
W-W'=ma
where a is the acceleration of the elevator. If we solve for a, we find
a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.
4 0
3 years ago
An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul
PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight  due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ =  400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

= 100 N .

weight at height = 100 N

5 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
Marizza181 [45]

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>
7 0
2 years ago
a train traveling at 30m/s comes to a complete stop in 20 sec.how far did the train go during the deceleration
elixir [45]

Answer:

600m

Explanation:

v=30m/s

t=20s

s=VT

20×30=600m

3 0
3 years ago
Which statement is true of earths poles?
djverab [1.8K]
Answer choice d is correct
7 0
3 years ago
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