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bija089 [108]
3 years ago
12

In a plate glass factory, sheets of glass move along a conveyor belt at a speed of 15.0 cm/s. An automatic cutting tool descends

at preset intervals to cut the glass to size. Since the assembly belt must keep moving at constant speed, the cutter is set to cut at an angle to compensate for the motion of the glass. If the glass is 72.0 cm wide and the cutter moves across the width at a speed of 24.0 cm/s, at what angle should the cutter be set
Physics
1 answer:
Leni [432]3 years ago
3 0

Answer:

angle at which the cutter should be set = 32°

Explanation:

We are given;

Speed of conveyor belt; v1 = 15 cm/s

Width;w = 72 cm

Speed of the cutter that moves across the width; v2 = 24 cm/s

Let's calculate time;

Since the width and the rate at which the cutter works per seconds is given, thus;

Time = w/v2

Time = 72/24 = 3 seconds

Now, we know that, velocity = distance/time

Thus, distance = velocity x time

So, distance covered;d = 15 x 3 = 45 cm

Now, for us to calculate the angle the cutter should be set, we would treat it as a triangle.

When the cutter cuts it, it covers a distance d which is perpendicular to the width.

Thus, to calculate the angle made with the width, we'll use trigonometric ratio.

Thus, d/w = tan θ where θ is the angle at which the cutter should be set.

So, 45/72 = tan θ

tan θ = 0.625

θ = tan^(-1)0.625

θ = 32°

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Answer:

see explanation

Explanation:

You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:

See the attached table.

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r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min

From the right we have:

r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.

And this should be the correct answer. Watch your chart and replace if it's neccesary.

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g= \frac{GM}{r^2}
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In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

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An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =
goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of  reynold number for both objects

R_{1}=R_{2}

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\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}

Here, k_{1}=k_{2}

h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}

h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}

Put the value into the formula

h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}

h_{2}=20\ W/Km^2

Hence, The value of the average convection coefficient is 20 W/Km².

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