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3 years ago

6

A 215 N sign is supported by two ropes. One rope pulls up and to the right 1=29.5∘ above the horizontal with a tension 1 , and t

he other rope pulls up and to the left 2=44.5∘ above the horizontal with a tension 2 , as shown in the figure. Find the tensions 1 and 2 .

1 answer:

WINSTONCH [101]3 years ago

8 0

The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:

∑ <em>F</em> (horizontal) = <em>T₁ </em>cos(29.5°) - <em>T₂</em> cos(44.5°) = 0

(right is positive, left is negative)

∑ <em>F</em> (vertical) = <em>T₁ </em>sin(29.5°) + <em>T₂</em> sin(44.5°) - 215 N = 0

(up is positive, down is negative)

Solve the system of equations. I use elimination here:

• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):

sin(29.5°) (<em>T₁ </em>cos(29.5°) - <em>T₂</em> cos(44.5°)) = 0

cos(29.5°) (<em>T₁ </em>sin(29.5°) + <em>T₂</em> sin(44.5°) - 215 N) = 0

<em>T₁ </em>cos(29.5°) sin(29.5°) - <em>T₂</em> cos(44.5°) sin(29.5°) = 0

<em>T₁ </em>cos(29.5°) sin(29.5°) + <em>T₂</em> cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)

• Subtract the first equation from the second to eliminate <em>T₁</em> :

<em>T₂</em> cos(29.5°) sin(44.5°) - (- <em>T₂</em> cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

• Solve for <em>T₂</em> :

<em>T₂</em> (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

<em>T₂</em> sin(74.0°) = (215 N) cos(29.5°)

… … … (using the fact that sin(<em>x</em> + <em>y</em>) = sin(<em>x</em>) cos(<em>y</em>) + cos(<em>y</em>) sin(<em>x</em>))

<em>T₂</em> = (215 N) cos(29.5°) / sin(74.0°)

<em>T₂</em> ≈ 195 N

• Solve for <em>T₁</em> :

<em>T₁ </em>cos(29.5°) - <em>T₂</em> cos(44.5°) = 0

<em>T₁ </em>cos(29.5°) = <em>T₂</em> cos(44.5°)

<em>T₁ </em>= <em>T₂</em> cos(44.5°) / cos(29.5°)

<em>T₁</em> ≈ 160. N

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g 2. In a laboratory experiment on standing waves a string 3.0 ft long is attached to the prong of an electrically driven tuning

abruzzese [7]

Answer:

The tension in string will be "3.62 N".

Explanation:

The given values are:

Length of string:

l = 3 ft

or,

= 0.9144 m

frequency,

f = 60 Hz

Weight,

= 0.096 lb

or,

= 0.0435 kgm/s²

Now,

The mass will be:

= $\frac{0.0435}{9.8}$

= $0.0044 \ kg$

As we know,

⇒ $\lambda=\frac{2L}{n}$

On substituting the values, we get

⇒ $=\frac{2\times 0.9144}{4}$

⇒ $=0.4572 \ m$

or,

⇒ $v=f \lambda$

⇒ $=0.4572\times 60$

⇒ $=27.432 \ m/s$

Now,

⇒ $v=\sqrt{\frac{T}{\mu} }$

or,

⇒ $T=\frac{m}{l}\times v^2$

On putting the above given values, we get

⇒ $=\frac{0.0044}{0.9144}\times (27.432)^2$

⇒ $=\frac{752.51\times 0.0044}{0.9144}$

⇒ $=3.62 \ N$

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3 years ago

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3 years ago

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. (Use equations not the psychrometric chart) The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 17oC, r

Fantom [35]

Answer:

a) The specific humidity of air is $9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$ .

b) The specific humidity of air is 0.464.

c) The dew-point temperature is 12.665 ºC.

Explanation:

a) The temperature of atmospheric air is considered the dry-bulb temperature, whereas the temperature of entirely saturated air is the the wet-bulb temperature. Dry bulb pressure is the atmospheric air. First we need to find the specific humidity at wet bulb temperature ( $\omega_{wb}$ ), measured in kilograms of water per kilogram of dry air:

$\omega_{wb} = \frac{0.622\cdot P_{wb}}{P_{db}-P_{wb}}$ (Eq. 1)

Where:

$P_{wb}$ - Wet bulb pressure, measured in kilopascals.

$P_{db}$ - Dry bulb pressure, measured in kilopascals.

Wet bulb pressure is the saturation pressure of water evaluated at wet bulb temperature, while dry bulb pressure in the pressure presented on statement. If $P_{db} = 95\,kPa$ and $P_{wb} = 1.9591\,kPa$ , then the specific humidity at wet bulb temperature is:

$\omega_{wb} = \frac{0.622\cdot (1.9591\,kPa)}{95\,kPa-1.9591\,kPa}$

$\omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}$

Now we use the following equation to determine the dry bulb specific humidity ( $\omega_{db}$ ), measured in kilograms of water per kilogram of dry air:

$\omega_{db} = \frac{c_{p,a}\cdot (T_{wb}-T_{db})+\omega_{wb}\cdot h_{fg,wb}}{h_{g,db}-h_{f,wb}}$ (Eq. 2)

Where:

$c_{p,a}$ - Isobaric specific heat of air, measured in kilojoules per kilogram-Celsius.

$T_{wb}$ - Wet-bulb temperature, measured in Celsius.

$T_{db}$ - Dry-bulb temperature, measured in Celsius.

$\omega_{wb}$ - Wet-bulb specific humidity, measured in kilograms of water per kilogram of dry air.

$h_{fg,wb}$ - Wet-bulb specific enthalpy of vaporization of water, measured in kilojoules per kilogram.

$h_{g,db}$ - Dry-bulb specific enthalpy of saturated vapor, measured in kilojoules per kilogram.

$h_{f,wb}$ - Wet-bulb specific enthalpy of liquid vapor, measured in kilojoules per kilogram.

If we know that $T_{wb} = 17\,^{\circ}C$ , $T_{db} = 25\,^{\circ}C$ , $c_{p,a} = 1.005\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $\omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}$ , $h_{fg, wb} = 2460.6\,\frac{kJ}{kg}$ , $h_{g,db} = 2546.5\,\frac{kJ}{kg}$ and $h_{f,wb} = 71.355\,\frac{kJ}{kg}$ , the dry bulb specific humidity is:

$\omega_{db} = \frac{\left(1.005\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (17\,^{\circ}C-25\,^{\circ}C)+\left(0.0131\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot \left(2460.6\,\frac{kJ}{kg} \right)}{2546.5\,\frac{kJ}{kg}-71.355\,\frac{kJ}{kg} }$

$\omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$

The specific humidity of air is $9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$ .

b) Then, the relative humidity of air ( $\phi_{db}$ ), dimensionless, is obtained from this expression:

$\phi_{db} = \frac{\omega_{db}\cdot P_{db}}{(0.622+\omega_{db})\cdot P_{sat, db}}$ (Eq. 3)

Where $P_{sat, db}$ is the saturation pressure at dry-bulb temperature, measured in kilopascals.

If we know that $\omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}$ , $P_{db} = 95\,kPa$ and $P_{sat, db} = 3.1698\,kPa$ , the relative humidity of air is:

$\phi_{db} = \frac{\left(9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot (95\,kPa)}{\left(0.622+9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}\right)\cdot 3.1698\,kPa}$

$\phi_{db} = 0.464$

The specific humidity of air is 0.464.

c) The dew point temperature is the temperature at which water is condensated when air is cooled at constant pressure. That temperature is equivalent to the saturation temperature at vapor pressure ( $P_{v}$ ), measured in kilopascals:

$P_{v} = \phi_{db} \cdot P_{sat, db}$ (Eq. 4)

( $\phi_{db} = 0.464$ , $P_{sat, db} = 3.1698\,kPa$ )

$P_{v} = 0.464\cdot (3.1698\,kPa)$

$P_{v} = 1.4707\,kPa$

The saturation temperature at given vapor pressure is:

$T_{dp} = 12.665\,^{\circ}C$

The dew-point temperature is 12.665 ºC.

4 0

4 years ago

I know nothing help me please

maxonik [38]

Answer:

have you ever feel like a plastic bag

Explanation:

Do you ever feel like a plastic bag

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Do you know that there's still a chance for you

'Cause there's a spark in you?

You just gotta ignite the light and let it shine

Just own the night like the 4th of July

'Cause, baby, you're a firework

Come on, show 'em what you're worth

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Wanting to start again?

Do you ever feel, feel so paper-thin

Like a house of cards, one blow from caving in?

Do you ever feel already buried deep

Six feet under screams but no one seems to hear a thing

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You just gotta ignite the light and let it shine

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Maybe a reason why all the doors are closed

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Come on, show 'em what you're worth

Make 'em go, "Ah, ah, ah"

As you shoot across the sky

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Come on, let your colors burst

Make 'em go, "Ah, ah, ah"

You're gonna leave 'em all in awe, awe, awe

Boom, boom, boom

Even brighter than the moon, moon, moon

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3 years ago

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Vikentia [17]

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Ray B is the reflected ray: This is the ray leaves the surface of a reflecting surface. The angle formed between the reflected ray and the normal is called reflected angle

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