Answer:
I want to believe the program is to be written in java and i hope your question is complete. The code is in the explanation section below
Explanation:
import java.util.Date;
public interface Downloadable {
//abstract methods
public String getUrl();
public Date getLastDownloadDate();
}
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 air heats up when being compressed and transefers heat to the barrel.
Explanation:
When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.
In an adiabatic transformation:
For air k = 1.4
SO
SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.
After it was compressed the hot air will exchange heat with the barrel heating it up.
Explanation:
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Answer:
critical stress required is 18.92 MPa
Explanation:
given data
specific surface energy = 1.0 J/m²
modulus of elasticity = 225 GPa
internal crack of length = 0.8 mm
solution
we get here one half length of internal crack that is
2a = 0.8 mm
so a = 0.4 mm = 0.4 × m
so we get here critical stress that is
...............1
put here value we get
=
= 18923493.9151 N/m²
= 18.92 MPa
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:
b). False
Explanation:
Lumped body analysis :
Lumped body analysis states that some bodies during heat transfer process remains uniform at all times. The temperature of these bodies is a function of temperature only. Therefor the heat transfer analysis based on such idea is called lumped body analysis.
Biot number is a dimensionless number which governs the heat transfer rate for a lumped body. Biot number is defined as the ratio of the convection transfer at the surface of the body to the conduction inside the body. the temperature difference will be uniform only when the Biot number is nearly equal to zero.
The lumped body analysis assumes that there exists a uniform temperature distribution within the body. This means that the conduction heat resistance should be zero. Thus the lumped body analysis is exact when biot number is zero.
In general it is assume that for a lumped body analysis, Biot number 0.1
Therefore, the smaller the Biot number, the more exact is the lumped system analysis.