This statement is false. Increasing the two objects' mass (I'm guessing) will actually increase their gravitational force. This is because of the equation:
If the distance was increased, then the statement would be true, but since you are increasing mass, which is proportional to the Force of Gravity, you are in fact, increasing the gravitational force between the two objects.
Answer:
Answer below!!!!
Explanation:
Convection is the transfer of thermal energy by particles moving through a fluid.
Hope I Helped!!!
;)
Answer: 313920
Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.
Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.
The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.
The depth below the water surface of each strip is,
di = 8 − yi
and that in turn gives us the pressure on the strip,
Pi =ρgdi = 9810 (8−yi)
The area of each strip is,
Ai = 2√4− (yi) 2Δy
The hydrostatic force on each strip is,
Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy
The total force on the plate is found on the attached image.
I’m assuming we’re suppose to get some kind of graph but, Instantaneous speed is the speed that is happening right now. Like driving a car at 15k/h. The instantaneous speed of the car 15k/h. On the graph, at 5s. Wherever the line is, will tell you what the speed is.
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
