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AysviL [449]
3 years ago
12

The average sound intensity inside a busy restaurant is 3.10 10-5 W/m2. How much energy goes into each ear (area = 2.1 10-3 m2)

during a 1 hour meal?

Physics
2 answers:
larisa [96]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

vichka [17]3 years ago
6 0

Answer:

2.344×10⁻⁴ J

Explanation:

Energy: This can be defined as the ability or the capacity to do wok.

The S.I unit of energy is Joule(J).

From the question,

E = P×t.............. Equation 1

Where E = amount of energy that goes into the ear, P = power, t = time.

But,

P = I×A.............. Equation 2

Where I = sound intensity, A = cross sectional area.

Substitute equation 2 into equation 1

E = I×A×t................ Equation 3

Given: 3.1×10⁻⁵ W/m², A = 2.1×10⁻³ m², t = 1 hour = 1×60×60 =  3600 s

Substitute into equation 3

E = 3.1×10⁻⁵× 2.1×10⁻³× 3600

E = 2.344×10⁻⁴ J.

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Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

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Please HELP ME If the motion of an object changes, what must be true about the forces acting on that object?
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If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. ... Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object.

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
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Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

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3 years ago
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