PHYSICS COLLEGE

What sound frequency could a human detect

Answers

Answer 1

Answer:

Answer:

People can hear sounds at frequencies from about 20 Hz to 20,000 Hz,

Answer 2

Answer:

20 Hz up to 20,000 Hz

Brain pls

Related Questions

If I am given a total capacitence of two capacitors, their capacitence togather is 22 F. What capacitence would the individual capacitors have if they are connected in parallel or connected in series.
An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.
Where is information first stored in a human brain
What is effort arm don't say the answer of gogle
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3 , is stirred until its temperature is 500 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, i

According to the U.S. Green Building Council, what percentage of the world’s energy use and greenhouse gas emissions can be attributed to buildings?

Answers

According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions

Answer:

40%

Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.

For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8 cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .

Answers

Answer:

Velocity = v = 2.24m/s

Acceleration = a = 0.20m/s²

Explanation:

Please see attachment below.

Given

z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

Final answer:

The roller coaster's velocity and acceleration at t=4 seconds is 7.64 m/s and 0.57 m/s² respectively.

Explanation:

The question is about understanding kinematics in cylindrical coordinates to analyze the motion of a roller coaster car. First, we need to understand that in polar coordinates, θ is changing with time t. So, the velocity vector v will have two components, one in the θ direction (rθ') and another in the z direction (z'). Given θ = 0.3t, we differentiate θ with respect to time to get θ' or dθ/dt = 0.3 rad/sec. Then, the z component of the velocity can be calculated by differentiating the equation of motion in the z-direction, z = -8 cos(θ), with respect to time. This gives z' = 8(0.3)sin(0.3t). So, at t=4s, z' = 8(0.3)sin(1.2) = 1.89 m/s. We then calculate rθ' = r*dθ/dt = 25*0.3 = 7.5 m/s.

The magnitude of velocity can then be calculated using the Pythagorean theorem: √((rθ')² + (z')²) = √((7.5)² + (1.89)²) = 7.64 m/s .

In a similar way, we can find the acceleration components. Given that r=25 m and is constant, radial acceleration is zero ( ar = r*(θ')²). The tangential acceleration is at = r*θ'' = r*d²θ/dt² =0 m/s² and z'' = dz'/dt = 8*0.3²*cos(0.3t). So, at t = 4s, z'' = 8(0.09)cos(1.2) = 0.57 m/s². The magnitude of the acceleration is given by √((ar)² + (at)² +(z'')²) = √((0)² + (0)² +(0.57)²)= 0.57 m/s².

Learn more about Motion in Cylindrical Coordinates here:

brainly.com/question/32575072

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A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +x-direction at 5.35 m/s2. At t; = 3.60 s, the rocket's fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 because of friction. A very particular physics professor wants the average velocity for the entire trip of the toy car to be +6.50 m/s. In order to make this happen, the physics professor plans to push the car (immediately after it comes to rest by friction) with a constant velocity for 4.50 sec. What displacement must the physics professor give the car (immediately after it comes to rest by friction) in order for its average velocity to be +6.50 m/s for its entire trip (measured from the time the rocket is lit to the time the physics professor stops pushing the car)?

Answers

Answer:

What displacement must the physics professor give the car

= 12.91 METERS

Explanation:

Check the attached file for explanation

Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of the square is 2a. What is the electric potential at P, the center of the square?a. kQ/4a
b. kQ/a
c. zero volts
d. 2kQ/a
e. 4kQ /a

Answers

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

$l = 2a$

The distance of the center of the square from each of the corners is

$r = (2a)/(2)= a$

The potential electric at the center due to each cornet charge is

$V_1 = (kQ_1)/(r_1)$

$V_2 = (kQ_2)/(r_2)$

$V_3 = (kQ_3)/(r_3)$

$V_4 = (kQ_4)/(r_4)$

The total electric potential at the center of the given square is

$V = V_1+V_2+V_3+V_4$

$V = (kQ_1)/(r_1)+ (kQ_2)/(r_2)+(kQ_3)/(r_3)+(kQ_4)/(r_4)$

Al the charges are equal, and the distance are equal to a, then

$V = (kQ)/(a)+ (kQ)/(a)+(kQ)/(a)+(kQ)/(a)$

$V = (4kQ)/(a)$

Therefore the correct option is E.

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

Answers

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) $h*(1 - 1/2 g * h/v_0^2)$

b) $h = v_0^2/ g$

c) $h = v_0^2/ g$

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

$y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t$

where:

y = height at time t

y0 = initial height

v0 = initial velocity

a = acceleration

t = time

v = velocity

a) When the balls collide, h1 = h2. Then,

$h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0$

Replacing in the equation of the height of the first ball:

$h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)$

b) that the balls collide at t = h/v0. Then:

$h/ v_0 = -v_0/-g\nh = v_0^2/ g$

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

$h = v_0^2/ g$

See more about velocity at brainly.com/question/862972

A boat can travel in still water at 56 m/s. If the boats sails directly across a river that flows at 126 m/s. What is the boats speed relative to the ground

Answers

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

Other Questions

. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.You should obtain e/m = 2V/(B^2)(r^2)3. The magnetic field on the axis of a circular current loop a distance z away is given byB = mu I R^2 / 2(R^2 + z^2)^ (3/2)where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/Rwhere B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake

A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.