A heat source transfers 3000 J/sec to a metal part surface. The heated area is circular, and the heat intensity decreases as the
1 answer:
Answer:
The resulting power density is enough to melt the metal.
Step-by-step explanation:
Given data:
Power = P = 3000 J/sec
diameter = d = 3.5 mm
Solution:
As we Know that Area = A = π r² ---- (1)
where r is radius.
Also Radius =
Putting the values of radius, π = 3.14 in equation 1, we get
A = 3.14 x ()²
A = 3.14 x (3.5/2)²
A = 9.62 mm²
As 75% of heat is concentrated in circular area then Power P becomes
P = 3000 J/sec x 75 %
As J/sec = Watt = W and 75 % = 3/4
so P = 3000 W x 3/4
P = 2250 W
As power density is represented by the formula:
Power density = PD = P/A
where P is Power and A is area.
So,
PD = P/A
Putting the values of Power and Area in above equation, we get
PD = 2250 W / 9.62 mm²
PD = 234 W/mm²
So, this power density is sufficient to melt the metal.
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Answer:
The equation contains exact roots at x = -4 and x = -1.
See attached image for the graph.
Step-by-step explanation:
We start by noticing that the expression on the left of the equal sign is a quadratic with leading term , which means that its graph shows branches going up. Therefore:
1) if its vertex is ON the x axis, there would be one solution (root) to the equation.
2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.
3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.
So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:
We recall that the x-position of the vertex for a quadratic function of the form is given by the expression:
Since in our case and
, we get that the x-position of the vertex is:
Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:
This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).
We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.
When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):
The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.
Then, the two roots are: x = -1 and x = -4.
Answer:
OR
Step-by-step explanation:
Hi there!
Point-slope form: where <em>m</em> is the slope of the line and
is a given point
Given that the slope is -3/4, we can plug it into as <em>m</em>:
We can also plug in the given point (1,2):
Slope-intercept form: where <em>m</em> is the slope and <em>b</em> is the y-intercept (the value of y when the line crosses the y-axis)
To write the equation in slope-intercept form, isolate <em>y</em>:
I hope this helps!