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sesenic [268]
3 years ago
14

An automobile tire is filled with air at a pressure of 27.0 lb/in2 at 25°C. A cold front moves through and the temperature drops

to 5°C. Assuming no change in volume, what is the new tire pressure?
Physics
1 answer:
Softa [21]3 years ago
7 0

Answer:

P_2=25.18\ lb/in^2

Explanation:

Given that

P₁=27 lb/in² ,T₁=25⁰C=273+25 = 298 K

T₂=5⁰C = 273 + 5  = 278 K

Lets take final pressure = P₂

We know that if volume of the gas is constant then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

P_2=P_1\times \dfrac{T_2}{T_1}

Now by putting the value in the above equation

P_2=27\times \dfrac{278}{298}\ lb/in^2

P_2=25.18\ lb/in^2

Therefore the new tire pressure will be

P_2=25.18\ lb/in^2

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Answer:

v =2.02

Explanation:

v^2=0.05-4.9

v^2=-4.85

square root both side

v=2.02

^^^^this is a not a perfect square  

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3 years ago
You fire a paintball from a paintball gun straight up into the air at 25.0 m/s. What is the ball velocity at the higher point th
iVinArrow [24]

Given :

You fire a paintball from a paintball gun straight up into the air at 25.0 m/s.

To Find :

What is the ball velocity at the higher point that the ball reaches.

Solution :

We know, when ball is moving upward acceleration due to gravity is applied downward.

So, the ball will decelerate and when ball reaches maximum height its velocity will become zero.

Therefore, ball velocity is 0 m/s at maximum height.

Hence, this is the required solution.

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3 years ago
James observes that the Polaris star in the northern hemisphere does not rise and set in the sky. His teacher tells him that thi
Nimfa-mama [501]

Answer:

Just above the pole (top-most red circle)

Explanation:

Polaris is used to identify North direction. Since, the Earth rotates on its axis which is along North-south, Polaris never seems to rise and set from the Northern hemisphere. This is because Polaris lies above north pole. Thus, in the given diagram, Polaris is above the North pole on the axis represented by top-most red circle.

7 0
3 years ago
Read 2 more answers
A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength eq
Veronika [31]
F = kq1q2/r<span>2
Where,
F - Coulomb Force
k - constant value which is equal to </span>8.98 × 10^9<span> newton square metre per square coulomb
q1 and q2 -  two electric charges
r - distance.

5.8 * 10^5 = 1.5 * 10^-9 / r^2
</span><span>5.8 * 10^5 r^2 = 1.5*10^-9
</span>r^2 = 0.0000258620
r = 0.0050854694

So the distance is equal to 5.09 x 10^-3 
3 0
3 years ago
O professor Hosney, levou os alunos da segunda série ao laboratório para realizar um experimento. Pegou um recipiente de capacid
givi [52]

Answer:

The temperature beyond which the substance overflows the container is 86.23°C.

Explanation:

English Translation

Professor Hosney took the second grade students to the laboratory to perform an experiment. He took a 1000ml capacity container at a temperature of 68oF and poured 980 ml of a substance at 20oC into it. While placing the set to heat, he consulted a table where he found the volumetric expansion coefficient of the substance, 4 x 10-4 ºC-1 and the linear expansion coefficient of the container material, 3 x 10-5 ºC-1. Hosney then asked students to determine the temperature from which the substance would overflow. A student then asked, what is the melting temperature of the substance, and the teacher answered promptly 290.8 K. What is the temperature from which the substance will overflow?

Solution

The change in volume of a substance is given as

ΔV = γV₀(ΔT)

where

γ = coefficient of volume expansion

V₀ = Initial volume

(ΔT) = change in temperature.

At the temperature where the substance will overflow, the volume of the substance and the container will both be the same.

Let this temperature be T.

For the substance,

γ = coefficient of volume expansion = (4 × 10⁻⁴) °C⁻¹

V₀ = Initial volume = 980 mL

(ΔT) = change in temperature = (T - 20)

We will still leave ΔT as ΔT

ΔV₁ = (4 × 10⁻⁴) × 980 × ΔT

ΔV₁ = 0.392 ΔT

New volume of the substance at that temperature = V₀ + ΔV₁ = 980 + 0.392ΔT

For the container

γ = coefficient of volume expansion = 3 × coefficient of linear expansion = 3 × (3 × 10⁻⁵) °C⁻¹ = (9 × 10⁻⁵) °C⁻¹

V₀ = Initial volume = 1000 mL

(ΔT) = change in temperature = (T - 20) (note that 68°F = 20°C)

We will still leave ΔT as ΔT

ΔV₂ = (9 × 10⁻⁵) × 1000 × ΔT

ΔV₂ = 0.09 ΔT

New volume of the container at that temperature = V₀ + ΔV₂ = 1000 + 0.09 ΔT

At the temperature where overflow occurs, the two volumes are initially first the same.

980 + 0.392ΔT = 1000 + 0.09 ΔT

0.392ΔT - 0.09ΔT = 1000 - 980

0.302ΔT = 20

ΔT = (20/0.302) = 66.23°C

T - 20° = 66.23°

T = 66.23 + 20 = 86.23°C

The temperature beyond which the substance overflows the container is 86.23°C.

Hope this Helps!!!

8 0
3 years ago
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