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3 years ago

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A machine with a mechanical advantage of 10 is used to produce an output force of 250 Newton’s. What input force is applied to t

his machine

1 answer:

finlep [7]3 years ago

4 0

Answer:

que tue nesisitas

Explanation:

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A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber

Kipish [7]

Answer:

The correct answers are

(a) It decreases to 1/3 L

(ii) is (c) It is constant

Explanation:

to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem

The given variables are as follows

Initial volume V1 = 1L

V2 = Unknown

Initial Temperature T1 = 300K

let us assume that the balloon is perfectly elastic

At 300K the balloon is filled and it stretches to maintain 1 atmosphere

at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,

For (i) since we know that the pressure of the balloon is constant

by Charles Law V1/T1 =V2/T2

or V2 = (V1/T1)×T2 = $\frac{1L}{300K}$ × $100K$ = $\frac{1}{3}$ × L = L/3 hence the correct answer to (i) is 1/3L

8 0

2 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ $ln[\frac{P_{1} }{P_{2} }]$

= <em>mRT</em>₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵<em>pV</em> = <em>mRT</em>]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = Δ<em>U </em>+ <em>W</em>

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago

In models of magnetic and electric fields, why are field vectors depicted by arrows?

taurus [48]

The arrows in models of magnetic and electric fields show both their magnitude and direction.

In Physics, a vector refers to a quantity that has both magnitude and direction. Hence, a vector always points in a given direction. The direction in which the arrow points is the direction of the vector in space.

In models of magnetic and electric fields, field vectors depicted by arrows because they represent both their magnitude and direction. The length of the arrow shows magnitude.

Learn more: brainly.com/question/102477

7 0

2 years ago

A closed cylinder with a 0.15-m radius ends is in a uniform electric field of 300 n/c, perpendicular to the ends. the total flux

bixtya [17]

The total flux through the cylinder is zero.

In fact, the electric flux through a surface (for a uniform electric field) is given by:

$\Phi = E A \cos \theta$

where

E is the intensity of the electric field

A is the surface

$\theta$ is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.

We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because $\theta=90^{\circ}$ and $\cos \theta=0$ ).

On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

$\Phi_1 = E \pi r^2$

where r is the radius, and where we have taken $\theta=0^{\circ}$ since the perpendicular to the surface is parallel to the direction of the electric field, so $\cos \theta=1$ . On the second surface, however, the perpendicular to the surface is opposite to the electric field, so $\theta=180^{\circ}$ and $\cos \theta=-1$ , therefore the flux is

$\Phi_2 = -E \pi r^2$

And the net flux through the cylinder is

$\Phi = \Phi_1 + \Phi_2 = E \pi r^2 - E \pi r^2=0$

4 0

3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago