Answer:
x = 727.5 km
Explanation:
With the conditions given using trigonometry, we can find the tangent
tan θ = CO / CA
With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun
D =150 10⁶ km (1000m / 1 km)
D = 150 10⁹ m.
We must take the given angle to radians.
1º = 3600 arc s
π rad = 180º
θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =
θ = 4.85 10⁻⁶ rad
That angle is extremely small, so we can approximate the tangent to the angle
θ = x / D
x = θ D
x = 4.85 10-6 150 109
x = 727.5 103 m
x = 727.5 km
Weight = (mass) x (gravity)
70 N = (mass) x (9.8 m/s²)
Divide each side by (9.8 m/s²) , and you have
mass = 70 N / 9.8 m/s² = 7.14 kg.
___________________________
Mass on the moon:
Mass doesn't change. It's a number that belongs to the bowling ball,
no matter where the ball goes. If the mass of the bowling ball is 7.14 kg
anywhere, then it's 7.14 kg everywhere ... on Earth, on the moon, on Mars, rolling around in the trunk of my car, or floating in intergalactic space.
However, WEIGHT depends on the gravity wherever the ball happens to be
at the moment.
The acceleration of gravity on the moon is 1.622 m/s².
So the WEIGHT of the ball on the moon is
(7.14 kg) x (1.622 m/s²) = 11.58 Newtons
That's only about 16% of its weight on Earth.
<u><em>Both solvent and solute are parts of a solution.</em></u> Solutions are mixtures of two or more substances, and the substance that dissolves into the solution is a solute. Meanwhile, the solute dissolves into a s Meanwhile, the solute dissolves into a substance called the solvent.
Solutes and solvents are substances not used only in chemical laboratories, but they are part of the day to day life. A solution contains only two components, which are solute and solvent. Solvent has the capability of dissolving the solute in a homogenous solution.
1) See attached figure
The relationship between charge and current is:
where
i is the current
Q is the charge
t is the time
Therefore, the current is the rate of change of the charge passing through a given point over time.
This means that for a graph of charge over time, the current is just equal to the slope of the graph.
For the graph in this problem:
- Between t = 0 and t = 2 s, the slope is
therefore the current is
i = 25 A
- Between t = 2 s and t = 6 s, the slope is
therefore the current is
i = -25 A
- Between t = 6 s and t = 8 s, the slope is
therefore the current is
i = 25 A
The figure attached show these values plotted on a graph.
2)
The previous equation can be rewritten as
This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.
Here we have the current vs time graph, so we gave to find the area under it.
The area of the first triangle is:
While the area of the second square is
So, the total area (and the total charge) is