PHYSICS COLLEGE

Since the density of air decreases with an increase in temperature, but the bulk modulus B is nearly independent of temperature, how would you expect the speed of sound waves in air to vary with temperature?

Answers

Answer 1

Answer:

To develop this problem it is necessary to apply the concept related to the speed of sound waves in fluids.

By definition we know that the speed would be given by

$v=\sqrt{(\beta)/(\rho)}$

$\beta =$ Bulk modulus

$\rho =$ Density of air

From the expression shown above we can realize that the speed of sound is inversely proportional to the fluid in which it is found, in this case the air. When the density increases, the speed of sound decreases and vice versa.

According to the statement then, if the density of the air decreases due to an increase in temperature, we can conclude that the speed of sound increases when the temperature increases. They are directly proportional.

Related Questions

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?
Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).
Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of the car applies a torque of 456 N·m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?
The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes the amplitude of damped oscillatory motion is _______. quadratic sinusoidal inverse exponential linear
Select the correct answer.Which statement is true if the refractive index of medium A is greater than that of medium B? A. O B. Total internal reflection is possible when light travels from air to medium B to medium A. Total internal reflection is possible when light travels from medium A to medium B. Total internal reflection is possible when light travels from medium B to medium A. D. Total internal reflection is possible when light travels from air to medium A. E. Total internal reflection is possible when light travels from air to medium B. C C.

Someone help Match the definition to the term.

1.

the noun that follows the verb and answers the questions

whom or what

linking verb

2. a word that expresses action

sentence

3. who or what the sentence is about

predicate noun

4.

a noun that indicates to or for whom or what something is

done

predicate adjective

5. follows a linking verb and renames the subject

direct object

subject

6. follows a linking verb and describes a subject

verb that joins a subject and a predicate noun or predicate

7.

adjective

indirect object

8. expresses a complete thought

verb

Answers

Answer:

Explanation:

No 1 is linked to the option provided at no 5.

A direct object is the noun that follows the verb and answers the question.

No 2 is linked to the option provided at no 8.

A verb is a word that expresses action.

No 3 is linked to the option provided at no 5.

The subject is what or whom the sentence is about.

No 5 is linked to the option provided at no 3.

A predicate noun follows a linking verb, and renames the subject.

No 6 is linked to the option provided at no 4.

A predicate adjective follows a linking verb, and describes a subject.

No 7 is linked to the option provided at no 1.

A linking verb joins a subject and a predicate.

No 8 is linked to the option provided at no 2.

A sentence expresses a complete thought....

Final answer:

The terms like direct object, verb, subject, indirect object, predicate noun, predicate adjective, and sentence are explained in the context of English grammar and matched to their definitions.

Explanation:

Here are the correct matches for the terms:

The noun that follows the verb and answers the questions whom or what is typically referred to as the direct object.
A word that expresses action is a verb.
Who or what the sentence is about is known as the subject.
A noun that indicates to or for whom or what something is done is an indirect object.
Follows a linking verb and renames the subject is a predicate noun.
Follows a linking verb and describes a subject is a predicate adjective.
Expresses a complete thought is the definition for a sentence in the context of English grammar.

Learn more about English Grammar Terms here:

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The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

Answers

Answer:

The correct answer is "21195 N".

Explanation:

The given values are:

Tensile strength,

= 3000 MN/m²

Diameter,

= 3.0 mm

i.e.,

= 3×10⁻³ m

Now,

The maximum load will be:

= $Tensile \ strength* Area$

On substituting the values, we get

= $(3000* 10^6)((\pi)/(4) (3* 10^(-3))^2)$

= $(3000* 10^6)((3.14)/(4) (3* 10^(-3))^2)$

= $21195 \ N$

Final answer:

The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.

Explanation:

The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.

We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.

Learn more about Tensile Strength here:

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Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horizontally with a constant speed of 3.80 m/s relative to observer 2 standing on the sidewalk.a) What is the speed of the ball 1.00 s after it is released, as measured by observer 2?

b) What is the direction of travel of the ball 1.00 s after it is released, as measured relative to the horizontal by observer 2?

Answers

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

$v_x = 3.80 m/s$

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

$v_y = u_y + gt$

where

$u_y =0$ is the initial vertical speed

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Therefore, after t = 1.00 s, the vertical speed is

$v_y = 0 + (9.8)(1.00)=9.8 m/s$

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

$v=√(v_x^2 +v_y^2)=√((3.8)^2+(9.8)^2)=10.5 m/s$

b) $\theta = -68.8^(\circ)$

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

$\theta = tan^(-1) ((v_y)/(v_x))=tan^(-1) ((-9.8)/(3.8))=-68.8^(\circ)$

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so $v_x$ has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so $v_y$ is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

$\theta = -68.8^(\circ)$

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up with your arms flat against your sides, and when you are standing straight up holding your arms straight out to your sides. Estimate the ratio of the moment of inertia with your arms straight out to the moment of inertia with your arms flat against your sides. (Assume that the mass of an average adult male is about 80 kg, and that we can model his body when he is standing straight up with his arms at his sides as a cylinder. From experience in men's clothing stores, a man's average waist circumference seems to be about 34 inches, and the average chest circumference about 42 inches, from which an average circumference can be calculated. We'll also assume that about 20% of your body's mass is in your two arms, and that each has a length L = 1 m, so that each arm has a mass of about m = 8 kg.)

Answers

Answer:

I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

d = (d₁ + d₂) / 2

d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

M = M_man - 0.2 M_man = 80 (1-0.2)

M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis: Ic = ½ M r²

A rod that rotates at the end: I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

I₁ = 8.258 + 5.33 + 4.129

I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

I₂ = ½ M r² + 2 m D²

I₂ = ½ 64 0.508² + 2 8 0.508²

I₂ = 8,258 + 4,129

I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

I₁ / I₂ = 17,717 /12,387

I₁ / I₂ = 1.43

The normal is a line perpendicular to the reflecting surface at the point of incidence.

Answers

Answer:

True

Explanation:

The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.

The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.

Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.

Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?

Answers

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_(max)=(hc)/(\lambda)-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=(hc)/(\lambda)-K_(max)\nW=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(495*10^(-9)m)-0.5eV\nW=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_(max)=(hc)/(\lambda)\n\lambda=(hc)/(W+K_(max))\n\lambda=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(2.01eV+1.5eV)\n\lambda=3.54*10^(-7)m=354*10^(-9)m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

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Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubber square of the same size. The rubber dielectric is 2.5 mm thick, hasr2.5, andbreakdown field strength of 25 megavolts per meter. Find the voltage rating of the capacitor using a safetyfactor of 10.

The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which data set represents constant acceleration?