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drek231 [11]
3 years ago
5

Example of third class Lever first class Lever and second class Lever​

Physics
2 answers:
horsena [70]3 years ago
7 0

pregunta: como palancas ??

zlopas [31]3 years ago
4 0

Answer:

wrench wheel barrow tweezers

Explanation:

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What is the average velocity of the object between 14 and 22 seconds?
Slav-nsk [51]

-1.1m/s

Explanation:

Work out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V). If you have U, A and T, use V = U + AT. If you have S, U and T, use V = 2(S/T) - U.

4 0
3 years ago
A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder
QveST [7]

Answer:

\Delta \theta = 47.57^{\circ} C

Explanation:

given,

moles of air compressed, n = 1.70 mol

initial temperature, T₁ = 390 K

Power supply by the compressor, P = 7.5 kW

Heat removed = 1.3 kW

Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.

Time of compression = time of the hay revolution

             =\dfrac{1}{2}\ T

             =\dfrac{1}{2}\times \dfrac{1}{f}

             =\dfrac{1}{2}\times \dfrac{1}{1.833}

             =0.273 s

Using first law of thermodynamics

U = Q - W

now,

\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}

Power supplied \dfrac{\Delta W}{\Delta t} = 7.5 kW

heat removed \dfrac{\Delta Q}{\Delta t} = 1.3 kW

now,

\dfrac{\Delta U}{\Delta t} = 7.5 -1.3

\dfrac{\Delta U}{\Delta t} = 6.2 kW

we know,

\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}

 C_v for air = 5 cal/° mol

                   = 5 x 4.186 J/mol°C  = 20.93 J/mol°C

now,

\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}

\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}

\Delta \theta = 47.57^{\circ} C

the temperature change per compression stroke is equal to 47.57°C.

4 0
3 years ago
Indica si las siguientes acciones son el resultado de un efecto estatico o dinamico de una fuerza.
Oduvanchick [21]

Answer:

Sabemos por definición que la fuerza es lo que es capaz de producir cambios o deformaciones en un cuerpo y a esto se denomina efecto estático y cuando la alteración ejerce movimiento o reposo se denomina efecto dinámico, con esta base resolvemos que:

-estirar un muelle: efecto estatico

- devolver una volea: efecto dinámico

- aplastar la plastilina: efecto estático

- empujar el carro del supermercado: efecto dinámico

- inflar un globo: efecto estático

8 0
3 years ago
A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a
marshall27 [118]

Explanation:

The given data is as follows.

     mass (m) = 0.160 kg,            spring constant (k) = 8 n/m,

     Maximum speed (v_{m}) = 0.350 m/s

Formula for angular frequency is as follows.

          \omega = \sqrt{\frac{{k}{m}}

    \omega = \sqrt{\frac{{8}{0.160}}

    \omega = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

            \nu_{max} = A \omega

                  A = \frac{\nu}{\omega}

                      = \frac{0.35}{7.07}

                      = 0.05 m

Hence, value of amplitude is 0.05 m.

(b)   Displacement = 0.030 m

Formula for mechanical energy is as follows.

            M.E = \frac{1}{2}kA^{2}

Putting the values into the above formula as follows.

            M.E = \frac{1}{2}kA^{2}

                   = \frac{1}{2} \times 8 \times (0.05)^{2}

                   = 9.8 \times 10^{-3} Joule

For x = 0.03,

As,     P.E = \frac{1}{2} \times kx^{2}

                = \frac{1}{2} \times 8 \times (0.03)

                = 3.6 \times 10^{-3}

Hence, calculate the kinetic energy as follows.

            K.E = M.E - P.E

                  = (9.8 \times 10^{-3} - 3.6 \times 10^{-3}) J

                  = 6.2 \times 10^{-3} J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is 6.2 \times 10^{-3} J.

7 0
3 years ago
A weightlifter raises a 200-kg barbell with an acceleration of 3 m/s^2. How much force does the weightlifter use to raise the ba
lesya692 [45]

F = mass x acceleration

We have mass = 200kg

and acceleration = 3 m/s^2 so...

F = (200)(3)

F = 600 N

4 0
3 years ago
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