Answer:
The answer is: measuring instrument
Explanation:
The researcher develops test to measure the variables she is interested in. So, they are her measuring instrument.
The correct answer is
<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.
In fact, the ideal gas law can be written as
</span>

<span>where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can see that if the temperature T increases, then the term on the right in the equation increases, therefore the term on the left should increase as well. In order for this to be possible, at least one between p and V should increase, or also both of them. Therefore, the correct answer is C.</span>
The complete question is;
The gravitational force between two objects is ______ proportional to the products of the masses and _______ proportional to the square of the distance between them.
Answer:
First dash is directly
Second dash is inversely
Explanation:
From Newton's law of gravitation the gravitational force is given by the equation;
F = GMm/r²
Where:
F is gravitational force
M and m are two masses representing the two objects
r is the distance between them
G = universal gravitational constant
Thus, if we are to write it in proportion format, we will have;
F ∝ Mm/r²
From this proportion relationship, we can see that the Gravitational force is directly proportional to the products of the two masses and inversely proportional to the square of the distance between them.
Answer:
The location of the shear center o is 0.033 or 33 m
Explanation:
Solution
Recall that,
The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12
= 30 * 10 ^ ⁻⁶ m⁴
Now,
The first moment of inertia is
Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)
= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²
Thus,
The shear flow is,
q = VQ/I
so,
P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶
P = (16.67 x - 83. 33 x²)
The shear force resisted by the shorter web becomes
Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x
Then,
We take the moment at a point A
∑Mₐ = 0
- ( p * e)- (Vw₂ * 0.3 ) = 0
e = 0.11 p * 0.3/p
which gives us 0.033 m
= 33 m
Therefore the location of the shear center o is 0.033 or 33 m
Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.
Using the following formula for linear-motion, the missing variable can be solved:
s = Vi * t + 1/2 (a * t^2)
Where: s = displacement = 29.7 m
Vi = initial velocity = 8.5 m/s
a = acceleration = 9.8
t = time = ?
Substituting:
29.7 = 8.5t + 1/2 (9.8*t^2)
29.7 = 8.5t + 4.9t^2
Dividing both sides by 4.9:
6.06 = 1.73t + t^2
t^2 + 1.73t - 6.06 = 0
(t - 1.74)(t + 3.48) = 0
t = 1.74s
From the above values, the correct answer is 1.74 seconds.