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3 years ago

Machines

Physics

2 answers:

eduard3 years ago

8 0

Answer:

MA=3.75

Explanation:Given that, E=75%, VR=no. of pulley=5

NOW,

E=(MA/VR)*100%

75%=(MA/5)*100%

MA=(75%*5)*100%

MA=375%/100%

∵MA=3.75

Ainat [17]3 years ago

3 0

Explanation:

Hey there!!

Here, Given is,

Efficiency = 75%

VR = no. of pulleys = 5

Now,

$effeciency = (\frac{ma}{vr} ) \times 100\%$

$75\% = \frac{ma}{5} \times 100\%$

100% ma = 75%×5

$ma = \frac{75\% \times5}{100\%}$

$ma = \frac{375\%}{100\%}$

Therefore, MA is 3.75.

Hope it helps...

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A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod

11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.

(a) Find X(t) in the form c • cos(w_o*t— α)

(b) Find the amplitude 3 Period of motion of the body 1

mass: m = 200g = 0.200 kg

displacement: ΔX = 20 cm = 0.20 m

Spring Constant: K = 9/0.20 = 45 N/m

IV: X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

= -5/12 = c_2

"radians Technically Unitless"

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

=2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

=π/6 s

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4 years ago

Please help me Calculate the total energy stored in the capacitor

Neko [114]

La fórmula: P = W/t ó W = P x t. donde:

P = potencia

W = trabajo

t = tiempo

Otra fórmula de potencia es: P= I x V

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Una fórmula muy importante que también hay que tener en cuenta es: V = q/C que indica que el voltaje es proporcional a la carga que hay en un condensador.

De la fórmula de potencia P= I x V y considerando que la corriente es constante (corriente continua), entonces la potencia es proporcional al voltaje. Si el voltaje aumenta en forma lineal, la potencia aumentará igual. Ver el siguiente diagrama.

Como la potencia varía en función del tiempo, no se puede aplicar la fórmula W = P x t, para calcular la energía transferida. Pero observando el gráfico, se ve que esta energía se puede determinar midiendo el área bajo la curva de la figura.

Energía Almacenada en un Condensador - Capacitor

El área bajo la curva es igual a la mitad de la potencia en el momento “t”, multiplicada por “t”.

Entonces: W = (P x t) / 2. Pero se sabe que P = V x I. Si se reemplaza esta última fórmula en la anterior se obtiene: W = (V x I x t) / 2, y como I x t = CV = Q, entonces para saber cuanta energía (W) hay en un condensador usamos una de las siguientes fórmulas:

W = (CV2/2) julios

W = (QV/2) julios

W = (Q2/2C) julios

, donde:

W = Trabajo (Energía) en julios

C = Capacidad en faradios

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3 years ago

Read 2 more answers

Why is velocity squared in kinetic energy and mass multiplied by 1/2?

maxonik [38]

Remember the definition of work done.
Work done is force(F) times displacement(x)
∴ W = F.Δx
According to Newton's 2nd law of motion,

F = ma
∴ W = ma.Δx ---- (i)
Using the kinematical equation v²-u² = 2ax,
aΔx = (v²-u²)/2

Plug this value in (i),

∴W = m[ $\frac{v^{2}-u^{2} }{2}$ ]
∴W = $\frac{mv^{2} }{2} - \frac{mu^{2} }{2}$

Which is nothing but change in kinetic energy.
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3 years ago

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic

Drupady [299]

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$

Explanation:

The density of mercury at 0 degrees is $d=1.36\cdot 10^4 kg/m^3$

Let's take 1 kg of mercury. Its volume at 0 degrees is

$V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3$

The formula to calculate the volumetric expansion of the mercury is:

$\Delta V= \alpha V \Delta T$

where

$\alpha=180\cdot 10^{-6} K^{-1}$ is the cubic expansivity of mercury

V is the initial volume

$\Delta T$ is the increase in temperature

In this part of the problem, $\Delta T=100 C-0 C=100 C=100 K$

So, the expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3$

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We can apply the same formula we used before, the only difference here is that the increase in temperature is

$\Delta T=22 C-0 C=22 C=22 K$

And the volumetric expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3$

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nadya68 [22]

Answer:

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Explanation:

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