A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1490 Hz. The bird-watcher, however, hears a frequency of 1505 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound

Answer:

true it all changes

Explanation:

________________________

Answer:

Push or pull of an object is considered a force. Push and pull come from the objects interacting with one another. Terms like stretch and squeeze can also be used to denote force.

In Physics, force is defined as:

The push or pull on an object with mass that causes it to change its velocity.

Force is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction. The direction towards which the force is applied is known as the direction of the force and the application of force is the point where force is applied.

Answer:

they were hunter gatherers

Explanation:

Answer:

Both sound and waves

Explanation:

Thank me later

Answer: 5.08 L.

Explation down below

Answer:

electromagnetic waves only

Explanation:

I just took the test, Hope it helps!

Answer:

A: Electromagnetic waves only

Explanation:

This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.

The minimum horizontal velocity required is "2.6 m/s".

First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.

[tex]h=v_it+\frac{1}{2}gt^2[/tex]

where,

h = height = 3 m

vi = initial vertical speed = 0 m/s

t = time interval = ?

g = acceleration due to gravity = 9.81 m/s²

therefore,

[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]

t = 0.78 s

Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:

[tex]s = vt\\\\v = \frac{s}{t}[/tex]

where,

s = horizontal distance = 2 m

t =0.78 s

v = minimum horizontal velocity = ?

Therefore,

[tex]v = \frac{2\ m}{0.78\ s}[/tex]

v = 2.6 m/s

Learn more about equations of motion here:  

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

Learn more about work here:

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Answer:

The correct answer is - 93.24×10^4 kg/m^3.

Explanation:

Given:

height of cylinder: 50.5 mm

diameter = 52.0

then radius will be diameter/2 = 52/2 = 26

Formula:

Density = mass/ volume

Volume = πr^2h

solution:

Now the volume of a cylinder is v = (22/7)×r^2×h

= 22/7×26×26×50.5

= 107261.59 mm^3  

Now volume in cubic meter V =10.7261 ×10^(-5) m^3

So density d = m/V = 1/(10.7261 ×10^(-5))

Or d = 93.24×10^4 kg/m^3

Answer:

Diagram (3).

Explanation:

N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A ([tex]F_{A} = -F_{B}[/tex]).

The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).

The law clearly states a Force of attraction; the two objects come towards each other.

Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.

Read more:

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Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          [tex]\frac{1}{f_1} = \frac{1}{q}[/tex]

           q = f₁

2) for an object located at p = 25 cm

            [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]

        [tex]\frac{1}{f_2}[/tex] = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = [tex]\frac{1}{f}[/tex]

where the focal length is in meters

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

Answer:

[tex]23^{\circ}[/tex]

Explanation:

n = Refractive index of air = 1

[tex]n_1[/tex] = Refractive index of contact lens = 1.6

[tex]n_2[/tex] = Refractive index of cornea = 1.4

[tex]n_3[/tex] = Refractive index of fluid = 1.3

From Snell's law

[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]

[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]

[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]

The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].

The angle is the light traveling in the fluid will be 23⁰. Light is traveling in a particular direction with an angle.

"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for a given set of media,

The given data in the problem is;

n is the refractive index of air = 1

n₁ is the refractive index of contact lens = 1.6

n₂ is the refractive index of cornea = 1.4

n₃ is the refractive index of fluid = 1.3

According to Snell's law. The formula for Snell's law is

[tex]\rm n sin30^0 = n_1 sin\theta \\\\ \theta = sin^{- 1}(\frac{1sin30^0}{1.6} )\\\\ \theta = 18.21 ^0[/tex]

For contact lenses;

[tex]\rm n_1sin\theta = n_2 sin\theta_1 \\\\ \theta_1 = sin^{-1}\frac{1.6 \times sin 18.21^0}{1.4} \\\\ \theta_1 =20.92 ^0[/tex]

For fluid;

[tex]n_2 sin\theta_1 = n_2 sin \theta_3\\\\ \theta_3 = sin^{-1}\frac{1.4 sin 20.92^0}{1.3} \\\\ \theta_3 = 22.62 ^ 0 =23^0[/tex]

Hence the angle is the light traveling in the fluid will be 23⁰.

To learn more about snell's law refer to the link;

https://brainly.com/question/10112549

Answer:

Fluid mechanics is considered one of the toughest subdisciplines within mechanical and aerospace engineering. It is unique from almost any other field an undergraduate engineer will encounter. It requires viewing physics in a new light, and that's not always an easy jump to make.

Answer:

B) It appears to be SHM (simple harmonic motion) which is also the projection of an object moving in a circular path on either the x or y axes.

Answer:

0.1 kg/s.

Explanation:

The density of water, d = 1000 kg/m³

Volume, V = 2 L

Time, t = 20 s

We need to find the mass flow rate from the faucet. We know that the density of an object is given by :

[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]

So, the mass flow rate is equal to 0.1 kg/s.

Answer:

I think its d

Explanation:

I'm not sure I'm sorry if I'm wrong

Answer:

   a = 7.29 m / s²,      T = 0.40 N

Explanation:

To solve this exercise we must apply Newton's second law to each body

The needle

              W -T = m a

              mg - T = ma

The spool, which we will approach by a cylinder

             Σ τ = I α

             T R = I α

the moment of inertia of a cylinder with an axis through its center is

             I = ½ M R²

angular and linear variables are related

            a = α R

            α = a / R

we substitute

           T R = (½ M R²) a / R

            T = ½ M a

we write our system of equations together

              mg - T = m a

                      T = ½ M a

we solve

              m g = (m + ½ M) a

              a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]

let's calculate

              a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]

              a = 7.29 m / s²

now we can look for the tension

              T = ½ M a

              T = ½ 0.110 7.29

               T = 0.40 N

Answer:

The answer is "15.56 cm".

Explanation:

[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]

Calculating object of length is x so:

[tex]u= -x[/tex]

Using formula:

[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

Answer:

[tex]4.25\ \text{m/s}[/tex]

[tex]3391.22\ \text{N}[/tex]

Explanation:

y = Height of compression = 0.38 m

m = Mass of basketball player = 101 kg

h = Height of center of gravity after jump = 0.92 m

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Energy balance of the system is given by

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]

The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]

[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]

The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

[tex]P=\dfrac{V^2}{R}[/tex]

Substitute all the values,

[tex]P=\dfrac{3^2}{2}\\\\P=4.5\ W[/tex]

So, the power of this circuit is equal to 4.5 watts.

Answer:

Ratio of kinetic energy of object A to B = 1:5

Explanation:

Given:

Mass of object A = 200 kg

Mass of object B = 1,000 kg

Find:

Ratio of kinetic energy of object A to B

Computation:

Kinetic energy = (1/2)(m)(v²)

Kinetic energy of object A = (1/2)(200)(v²)

Kinetic energy of object A = (100)(v²)

Kinetic energy of object B = (1/2)(1,000)(v²)

Kinetic energy of object B = (500)(v²)

Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B

Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)

Ratio of kinetic energy of object A to B = 100 / 500

Ratio of kinetic energy of object A to B = 1/5

Ratio of kinetic energy of object A to B = 1:5

Answer:

λ = a

Explanation:

This is a diffraction exercise that is described by the expression

          a sin θ = m λ

         sin θ  = m λ/ a

the first zero of the diffraction occurs for m = 1

        sin θ  = λ / a

angles are generally very small and are measured in radians

         sin θ  = θ  = y / x

we substitute

         [tex]\frac{y}{x} = \frac{\lambda}{a}[/tex]

the width of the central maximum is twice the distance to zero

         w = 2y

in the exercise indicate that this width is equal to twice the distance to the screen (2x)

          W = 2x

           2y = 2x

we substitute

          1 = λ/ a

          λ = a

we see that the width of the slit is equal to the wavelength used.