Physics, 21.10.2020 04:01 Shobhoob

Find the weight of a pumpkin hanging from two scales, if scale one reads 55 N, scale two reads 58 N, and the angle θ between the strings coming from the two scales is 121 degrees.

Find the weight of a pumpkin hanging from two scales, if scale one reads 55 N, scale two reads 58 N

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Answer from: juliopejfuhrf7541

The weight of the pumpkin hanging from the two scales is 98.36 N.

The given parameters;

reading on scale one, a = 55 Nreading on scale two, b = 58 Nangle between the two scales, θ = 121⁰

The weight of the pumpkin hanging from the two scales is equal to the resultant of the two scale readings.

The resultant of the two force is calculated as follows;

R² = a² + b² - 2abcos(θ)

R² = 55² + 58² - 2(55)(58)cos(121)

R² = 9674.7

$R = \sqrt{9674.7} \\\\R = 98.36 \ N$

Thus, the weight of the pumpkin hanging from the two scales is 98.36 N.

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Answer from: andrecoral105

The weight of the pumpkin is 46.793 newtons.

Explanation:

We consider the string-pumpkin system as our system, in which weight of pumpkin create two distinct tensions through cords. The equations of equilibrium are now described:

$\Sigma F_{x} = -T_{1}\cdot \sin \theta_{1}+T_{2}\cdot \sin \theta_{2} = 0$ (Eq. 1)

$\Sigma F_{y} = T_{1}\cdot \cos \theta_{1}+T_{2}\cdot \cos \theta_{2} -W = 0$ (Eq. 2)

Where:

$T_{1}$ , $T_{2}$ - Tension through cords, measured in Newtons.

- Weight of the pumpkin, measured in Newtons.

$\theta_{1}$ , $\theta_{2}$ - Inclination of each cord with respect to the vertical, measured in sexagesimal degrees.

If we know that $T_{1} = 55\,N$ , $T_{2} = 58\,N$ , $\theta_{1} = \theta$ and $\theta_{2} = 121^{\circ}-\theta$ , the system of equations becomes:

$-55\cdot \sin \theta +58\cdot \sin (121^{\circ}-\theta) = 0$ (Eq. 1b)

$55\cdot \cos \theta + 58\cdot \cos (121^{\circ}-\theta) = W$ (Eq. 2b)

From Trigonometry, we remember the following identity:

$\sin (121^{\circ}-\theta) = \sin 121^{\circ}\cdot \cos \theta -\cos 121^{\circ}\cdot \sin \theta$

$\sin (121^{\circ}-\theta) = 0.857\cdot \cos \theta -0.515\cdot \sin \theta$ (Eq. 3)

By applying (Eq. 3) in (Eq. 1b), we get the following expression after expanding and simplifying algebraically:

$-55\cdot \sin \theta + 58\cdot (0.857\cdot \cos \theta - 0.515\cdot \sin \theta) = 0$

$-84.87\cdot \sin \theta +49.706\cdot \cos \theta = 0$

And we solve the equation for $\theta$ :

$\tan \theta = \frac{49.706}{84.87}$

$\theta = \tan^{-1} 0.585$

$\theta \approx 30.328^{\circ}$

Then, the weight of the pumpkin is: ( $\theta \approx 30.328^{\circ}$ )

$W = 55\cdot \cos 30.328^{\circ}+58\cdot \cos (121^{\circ}-30.328^{\circ})$

$W = 46.793\,N$

The weight of the pumpkin is 46.793 newtons.

Answer from: Quest

as technology advances, new experiments often expose problems in accepted theories. c

hope i

-lavira: )

Answer from: Quest

the grams of oxygen gas that can be produced is 20.6 grams of o₂

calculation

2liclo₃ → 2 licl + 3 o₂

step 1: find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5 +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281 moles

step 2: use the mole ratio to calculate the moles of o2

from equation above, liclo₃: o₂ is 2: 3

therefore the moles of o₂=0.4281 moles x3/2 =0.6422 moles

step 3: find mass of o₂

mass= moles x molar mass

from periodic table the molar mass of o₂ = 16 x2= 32 g/mol

mass=0.6422 moles x 32 g/mol = 20.6 grams of o₂

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