Answer:
d=1.49×1011m
Explanation:
Velocity is defined as the rate of travel, and can be found using the distance formula.
velocity=distancetime
Rearranging this formula we can solve for distance given velocity and time of travel.
d=vt
We are given velocity and time, and so can solve for distance, but if we plug in the values given;
d=(3.00×108m/s)(8.3minutes)
We can see that the units do not match up. Since seconds are the SI unit for time, we will need to convert 8.3 minutes to seconds.
t=(8.3minutes)(60seconds/minute)=(498s)
Now our units work out and we can solve for distance.
= 15.85
Answer:
If I remembered correctly the answer would be Ne and Ar(Neon and Argon
If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.
The resultant of the two forces F1 and F2 is

with direction at

north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be

south-east.
Answer:
Q = 40.1 degrees
Explanation:
Given:
- The weight of the timber W = 670 N
- Water surface level from pivot y = 2.1 m
- The specific density of water Y = 9810 N / m^3
- Dimension of timber = (0.15 x 0.15 x 0.0036) m
Find:
- The angle of inclination Q that the timber makes with the horizontal.
Solution:
- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:
F_b = Y * V_timber
F_b = 9810*0.15*0.15*x
F_b = 226.7*x N
- Take static equilibrium conditions for the timber, and take moments about the pivot:
(M)_p = 0
W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0
- Plug values in:
670*0.5*3.6 - x^2 * 0.5*226.7 = 0
x^2 = 1206 / 113.35
x = 3.26 m
- Now use the value of x and vertical height y to compute the angle of inclination to be:
sin(Q) = y / x
sin(Q) = 2.1 / 3.26
Q = sin^-1 (0.6441718)
Q = 40.1 degrees
Sorry I didn't see this before...
Okay, I see two major problems with this student's experiment:
1) Nitric acid Won't Dissolve in Methane
Nitric acid is what's called a mineral acid. That means it is inorganic (it doesn't contain carbon) and dissolves in water.
Methane is an organic molecule (it contains carbon). It literally cannot dissolve nitric acid. Here's why:
For nitric acid (HNO3) to dissolve into a solvent, that solvent must be polar. It must have a charge to pull the positively charged Hydrogen off of the Oxygen. Methane has no charge, since its carbon and hydrogens have nearly perfect covalent bonds. Thus it cannot dissolve nitric acid. There will be no solution. That leads to the next problem:
2) He's Not actually Measuring a Solution
He's picking up the pH of the pure nitric acid. Since it didn't dissolve, what's left isn't a solution—it's like mixing oil and water. He has groups of methane and groups of nitric acid. Since methane is perfectly neutral (neither acid nor base), the electronic instrument is only picking up the extremely acidic nitric acid. There's no point to what he's doing.
Does that help?