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3 years ago

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An automobile tire having a temperature of 6.7 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 25 lb/in2 . What

would be the gauge pressure in the tire when its temperature rises to 33◦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 .

1 answer:

OlgaM077 [116]3 years ago

3 0

Answer: The gauge pressure in the tire when its temperature rises to 33◦C will be 28.7 lb/in2

Explanation: Please see the attachments below

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An electric fan is running on HIGH. After the LOW button is pressed, the angular speed of the fan decreases to 84.7 rad/s in 2.9

ale4655 [162]

The initial angular speed of the fan will be 55.0 rad/sec. The angular speed of the fan decreases to 84.7 rad/s in 2.96 s.

<h3>What is angular acceleration?</h3>

Angular acceleration is defined as the pace of change of angular velocity with reference to time.

Given data;

Final angular speed, $\rm \omega_f = 84.7 rad/s$

Initial angular speed, $\rm \omega_i = ?$

Time period,t= 2.96 s

Angular deceleration = 47.2 rad/s²

$\rm \alpha =\frac{\omega_f-\omega_i}{t} \\\\\rm 47.2 =\frac{84.7-\omega_i}{2.96} \\\\ \omega_i = 55.0 \ rad/sec$

Hence the initial angular speed of the fan will be 55.0 rad/sec.

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2 years ago

A 10kg object experiences a horizontal force which causes it to accelerate 5 m/s2 moving it a distance of 20 m horizontally. How

mr_godi [17]

Answer:

The work done by the force is 1000 J.

Explanation:

Given:

Mass of object (m) = 10 kg

Acceleration of the object (a) = 5 m/s²

Displacement of the object (S) = 20 m

Using Newton's second law, we have:

Force = Mass × Acceleration

$F=ma$

Plug in the given values and solve for force 'F' acting on the object. This gives,

$F=(10\ kg)(5\ m/s^2)\\\\F=50\ N$

Now, work done by the force 'F' is equal to the product of force 'F' and the displacement it cause in its direction which is 20 m.

Therefore, the work done by the force is given as:

Work = Force × Displacement

$W=FS$

Plug in the values given and solve for 'W'. This gives,

$W=50\ N\times 20\ m \\\\W=1000\ J$

Therefore, the work done by the force is 1000 J.

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4 years ago

slab of ice floats on water with a large portion submerged beneath the water surface. The slab is in the shape of a rectangular

n200080 [17]

Answer:

a) $\%V = 87.36\,\%$ , b) $x = 1.248\,m$ , c) $F_{B} = 176488.341\,N$ , d) Six polar bears.

Explanation:

a) The slab of ice is modelled by the Archimedes' Principles and the Newton's Laws, whose equation of equilibrium is:

$\Sigma F =\rho_{w}\cdot g \cdot A \cdot x-\rho_{i}\cdot g\cdot V = 0$

The height of the ice submerged is:

$\rho_{w}\cdot A \cdot x = \rho_{i}\cdot V$

$x = \frac{\rho_{i}\cdot V}{\rho_{w}\cdot A}$

$x = \frac{\left(900\,\frac{kg}{m^{3}}\right)\cdot (20\,m^{3})}{\left(1030\,\frac{kg}{m^{3}} \right)\cdot (14\,m^{2})}$

$x = 1.248\,m$

The percentage of the volume of the ice that is submerged is:

$\%V = \frac{(1.248\,m)\cdot (14\,m^{2})}{20\,m^{3}} \times 100\,\%$

$\%V = 87.36\,\%$

b) The height of the portion of the ice that is submerged is:

$x = 1.248\,m$

c) The buoyant force acting on the ice is:

$F_{B} = \left(1030\,\frac{kg}{m^{3}} \right)\cdot (1.248\,m)\cdot (14\,m^{2})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F_{B} = 176488.341\,N$

d) The new system is modelled after the Archimedes' Principle and Newton's Laws:

$\Sigma F = -n\cdot m_{bear}\cdot g-\rho_{i}\cdot V \cdot g + \rho_{w}\cdot V\cdot g = 0$

The number of polar bear is cleared in the equation:

$n\cdot m_{bear} = (\rho_{w} - \rho_{i})\cdot V$

$n = \frac{(\rho_{w}-\rho_{i})\cdot V}{m_{bear}}$

$n = \frac{\left(1030\,\frac{kg}{m^{3}} - 900\,\frac{kg}{m^{3}} \right)\cdot (20\,m^{3})}{400\,kg}$

$n = 6.5$

The maximum number of polar bears that slab could support is 6.

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3 years ago

Which of these discoveries is generally attributed to kepler?

Ne4ueva [31]

The question is incomplete and no options are given but the answer is;
The orbit of Mars is an ellipse,<span> is generally attributed to Kepler.
Kepler's first law states that; "</span><span>The orbits of the planets are ellipses, with the Sun at one focus of the ellipse.
</span>The planet at that point takes after and follow the ellipse in its orbit, which implies that the distance between Earth and Sun remove is continually changing as the planet goes around its orbit.

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3 years ago

You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6

marta [7]

Answer:

(a) The current needed is 56.92 A

(b) The magnitude of the magnetic field inside the solenoid is 0.134 T

(c) The energy density inside the solenoid is 7.144 kJ/m³

Explanation:

Given;

energy stored in the magnetic field of solenoid, E = 8.1 J

number of turns of the solenoid, N = 620 turns

diameter of the solenoid, D = 6.6 cm = 0.066 m

radius of the solenoid, r = D/2 = 0.033 m

length of the solenoid, L = 33 cm = 0.33 m

Inductance of the solenoid is given as;

$L= \frac{\mu_o N^2 A}{l}$

where;

A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

$L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H$

(A). How much current needed

Energy stored in magnetic field of solenoid is given as;

$E = \frac{1}{2} LI^2\\\\$

Where;

I is the current in the solenoid

$E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A$

(B) The magnitude of the magnetic field inside the solenoid

B = μ₀nI

where;

n is number of turns per unit length

B = μ₀(N/L)I

B = (4π x 10⁻⁷)(620/0.33)(56.92)

B = 0.134 T

(C) The energy density (energy/volume) inside the solenoid

$U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3$

3 0

3 years ago