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pogonyaev
3 years ago
14

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro

m a storage room to a machinist. Calculate the weight of the rod, w. The acceleration due to gravity, g = 9.81 m/s2 .
Physics
1 answer:
vichka [17]3 years ago
4 0

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

First, to calculate the weight of any object, we use the 2nd law of newton. In this case, the weight is:

w = m*g  (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V  

From here, we solve for mass:

m = d * V   (2)

Finally, we can know the volume of the rod, because is cylindrical, therefore, the volume of a cylinder is:

V = π * r² * h   (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

diameter = 2.3 cm ---> radius = 2.3/2 = 1.15 cm -----> 0.0115 m

Length or height = 88.8 cm ----> 0.888 m

Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

m = 2.88 kg

Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>
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So, based on the angle values that have been found, the angle of elevation of the nozzle can be <u>16° or 74°</u>.

<h3>Introduction</h3>

Hi ! This question can be solved using the principle of parabolic motion. Remember ! When the object is moving parabolic, the object has two points, namely the highest point (where the resultant velocity is 0 m/s in a very short time) and the farthest point (has the resultant velocity equal to the initial velocity). At the farthest distance, the object will move with the following equation :

\boxed{\sf{\bold{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}}}

With the following condition :

  • \sf{x_{max}} = the farthest distance of the parabolic movement (m)
  • \sf{v_0} = initial speed (m/s)
  • \sf{\theta} = elevation angle (°)
  • g = acceleration due to gravity (m/s²)

<h3>Problem Solving :</h3>

We know that :

  • \sf{x_{max}} = the farthest distance of the parabolic movement = 2.5 m
  • \sf{v_0} = initial speed = 6.8 m/s
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Step by Step :

  • Find the equation value \sf{\bold{theta}} (elevation angle)

\sf{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}

\sf{x_{max}  \cdot g = (v_0)^2 \cdot \sin(2 \theta)}

\sf{\frac{x_{max}  \cdot g}{(v_0)^2} =   \sin(2 \theta)}

\sf{\frac{2.5  \cdot 9.8}{(6.8)^2} =   \sin(2 \theta)}

\sf{\frac{2.5  \cdot 9.8}{(6.8)^2} =  \sin(2 \theta)}

\sf{\frac{24.5}{46.24} = \sin(2 \theta)}

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\sf{2 \theta = 32^o + k \cdot 360^o}

\sf{\theta = 16^o + k \cdot 180^o}

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→ \sf{k = 1 \rightarrow 16^o + 180^o = 196^o} (F)

Second Probability

\sf{2 \theta = (180^o - 32^o) + k \cdot 360^o}

\sf{2 \theta = 148^o + k \cdot 360^o}

\sf{\theta = 74^o + k \cdot 180^o}

→ \sf{k = 0 \rightarrow 74^o + 0 = 74^o} (T)

→ \sf{k = 1 \rightarrow 74^o + 180^o = 254^o} (F)

\boxed{\sf{\therefore \theta \{16^o , 74^o\} }}

<h3>Conclusion</h3>

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