All categories

3 years ago

True of false efficiency compared the output work to the output force

1 answer:

8 0

The statement about "efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.

You might be interested in

A solid block of mass m2 = 1.14 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring

borishaifa [10]

Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

initial momentum = final momentum

$m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v^{2}$

where

m1 = 1.14 kg

v1 = 2.0 m/s

m2 = 1.14 kg

v2 = 0 m/s

putting all value in the above equation

$1.14 *2.0+ 0 =(1.14+1.14)v^{2}$

$v =\frac{1.14*2.0}{1.14+1.14}$

v = 1 m/s

4 0

3 years ago

What Converts sound to a signal in a telephone

Irina-Kira [14]

The part you talk into, that converts the sound of your voice
into an electrical signal, is a tiny microphone.

-- The sound waves from your voice are ripples in the air.

-- In most microphones, there's a tiny coil of wire hanging
between the ends of a tiny magnet.

-- When the ripples in the air hit the little coil of wire, they
make it vibrate (wiggle) slightly.

-- When a coil of wire wiggles in the field of a magnet,
a current flows in the wire.

There's your electrical signal !

6 0

3 years ago

Which of the diagrams below illustrates sound waves generated by a siren

stepladder [879]

Diagram D. shows the sound waves generated by a siren

that is moving with constant speed to the left.

A sound wave is the sample of disturbance caused by the movement of strength journeying thru a medium because it propagates far away from the supply of the sound. Sound waves are created by using object vibrations and bring strain waves, for example, a ringing cellular phone.

Sound waves fall into three classes: longitudinal waves, mechanical waves, and strain waves. keep studying to find out what qualifies them as such. Longitudinal Sound Waves A longitudinal wave is a wave wherein the movement of the medium's debris is parallel to the course of the energy transport. Sound propagates via air or different mediums as a longitudinal wave, in which the mechanical vibration constituting the wave occurs along the direction of propagation of the wave.

Learn more about sound waves here:-brainly.com/question/1199084

#SPJ9

3 0

1 year ago

The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele

Alenkinab [10]

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

$P=P_0e^{-\frac{Mg}{RT}h}$

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.

Furthermore, the specific gas constant is defined by:

$R_{H_2}=\frac{R}{M}$

Therefore yo can write the barometric formula as:

$P=P_0e^{-\frac{g}{R_{H_2}T}h}$

at the surface of the planet (h =0) the pressure is $P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}$

applying to the previuos equation:

$\frac{P_0}{2} =P_0e^{-\frac{g}{R_{H_2}T}h}$

solving for h:

h=17357.9m

3 0

3 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago