Answer:
Explanation:
Adjusting the distance between the two electrodes is called gapping your spark plugs. You need a feeler gauge to gap your spark plugs properly If you're re-gapping a used plug, make sure that it's clean (gently scrub it with a wire brush)
hope this you
Answer:
the magnitude of F_A is 752 N
the direction theta of F_A is 57.9°
Explanations:
Given that,
Resultant force = 1330 N in x direction
∑Fx = R
from the diagram of the question which i uploaded along with this answer
FB = 800 N
FAsin∅ + FBcos30 = 1330 N
FAsin∅ = 1330 - (800 × cos30)
FA = 637.18 / sin∅
Now ∑Fx = 0
FAcos∅ - FBsin30 = 0
we substitute for FA
(637.18 / sin∅)cos∅ = 800 × sin30
637.18 / 800 × sin30 = sin∅/cos∅
and we know that { sin∅/cos∅ = tan∅)
so tan∅ = 1.59295
∅ = 57.88° ≈ 57.9°
THEREFORE FROM THE EQUATION
FA = 637.18 / sin∅
we substitute ∅
so FA = 637.18 / sin57.88
FA = 752 N
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.
Answer:
Explanation:
To convert to radians
A31∘43′53′′, 90∘32′11′′, 57∘43′56′′
using DMS approach ; 1degree = 60minutes = 3600 seconds
1° = 60' = 3600"
And degree to radian = multiply by π/180
A) 31∘43′53′′ = 31degree + 43minutes + 53 seconds
= 31 degree + 43minutes + 53/60
= 31 degree + 43.88minutes
= 31 degree + 43.88/60 = 31.73 degree x π/180 = 0.5534radians
FOR 90∘32′11′′ = 90 degree + 32minutes + 11seconds
= 90degree + 32minutes + 11/60
= 90 degree + 32.183minutes
= 90 degree + 32.183/60 = 90.54degree x π/180
= 1.580radians
FOR 57∘43′56′′ = 57degree + 43minutes+ 56seconds
= 57degree + 43minutes + 56/60
= 57 degree + 43.93minutes
= 57degree + 43.93/60 = 57.73degree X π/180
= 1.00radians
PART B
FOR 94∘22′19′′ = 94degree + 22minutes + 19seconds
= 94degree + 22minutes + 19/60
= 94degree + 22.32minutes
= 94degree + 22.32/60
= 94.37degree X π/180 = 1.65radians
FOR 40∘54′53′′ = 40degree + 54minutes + 53seconds
= 40 degree + 54minutes + 53/60
= 40 degree + 54.88minutes = 40 degree + 54.88/60
= 40.91degree X π/180 = 0.714radians
FOR 44∘42′48′′ = 44degree + 42minutes + 48seconds
= 44degree + 42.8minutes
= 44.71degree X π/180 = 0.780radians
Answer:
A.
0.176270π rad, 0.502980π rad, 0.320735π rad
B.
0.524289π rad, 0.227304π rad, 0.248407π rad
Explanation:
We know that,
1° = 60' 180° = π
1 ' = 1°/60 1° = π/180
A.
a. 31°43'53''
Step 1
53'' = 53 * 1/60
= 53'/60
Step 2
43'53''
= 43'+53'/60
= (2580+43)/60
= 2623'/60
-------- Convert to degrees
= 2623/60 * 1/60
= 2623/3600
Step 3
31°43'53''
= 31+ 2623/3600
= (111600 + 2623)/3600
= 114223°/3600
Now, we convert to radians
= 114223/3600 * π/180°
= 0.176270π rad
b.
90°32'11''
Step 1.
11' = 11 * 1/60
= 11/60
Step 2
32'11'
= 32 + 11/60
= 1931/60
-------- Convert to degrees
= 1931/60 * 1/60
= 1931/3600
Step 3
90°31'11''
= 90 + 1931/3600
= 325931°/3600
Now we convert to radians
= 325931°/3600 * π/180°
= 0.502980π rad
c.
57°43'56''
Step 1
56' = 56 * 1/60
= 56/60
= 14/15
Step 2
43'56''
= 43 + 14/15
= 659/15
Now we convert to degrees
= 659/15 * 1/60
= 659°/900
Step 3
57°43'56''
= 57 + 659/900
= 51959/900
Now we convert to radians
= 51959°/900 * π/180°
= 0.320735π rad
B.
a.
94∘22′19′′
Step 1
19'' = 19/60
Step 2
22'19''
= 22 + 19/60
= 1339/60
Now we convert to degrees
= 1339/60 * 1/60
= 1339°/3600
Step 3
94°22'19"
= 94 + 1339/3600
= 339739°/3600
Now we convert to radians
= 339739°/3600 * π/180
= 0.524289π rad
b.
40∘54′53′′
Step 1
53" = 53/60
Step 2
54'53"
= 54'+ 53/60
= 3293/60
Now we convert to degrees
= 3293/60 * 1/60
= 3293/3600
Step 3
40°54'53"
= 40 + 3293/3600
= 147293/3600
Now we convert to radians
= 147293/3600 * π/180
= 0.227304π rad
c.
44∘42′48′
Step 1
48' = 48/69
= 4/5
Step 2
42'48"
= 42 + 4/5
=214/5
Nowz we convert to degrees
= 214/5 * 1/60
= 107/150
Step 3
44°42'48"
= 44 + 107/150
= 6707/150
Now we convert to radians
= 6707/150 * π/180
= 0.248407π rad
Answer:
The solution code is as follows:
Explanation:
Firstly, we initialize a 10-elements array, myArray (Line 7) with no values.
Next, we create a for-loop (Line 10). Within the loop, we prompt user to enter an integer and assign the input value to the current element of myArray (Line 12-13).
again = "false";
result = 0;
a. choice
b. again
c. result
d. none of these are Boolean variables
Answer:
C
Explanation:
Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.