The factors that effect the electrical force between two objects are the charge of the objects and the distance between the objects.
Answer:
v2 = 65 m/s
the speed of the water leaving the nozzle is 65 m/s
Explanation:
Given;
Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle
Initial speed v1 = 0.65 m/s
diameter d1 = 3.0 cm
diameter (nozzle) d2 = 0.3 cm
The volumetric flow rates in both the hose and the nozzle are the same.
V1 = V2 ........1
Volumetric flow rate V = cross sectional area × speed of flow
V = Av
Area = (πd^2)/4
V = v(πd^2)/4 ....2
Substituting equation 2 to 1;
v1(πd1^2)/4 = v2(πd2^2)/4
v1d1^2 = v2d2^2
v2 = (v1d1^2)/d2^2
Substituting the given values;
v2 = (0.65 × 3^2)/0.3^2
v2 = 65 m/s
the speed of the water leaving the nozzle is 65 m/s
Scientists have determined the temperature near the Earth's center to be 6000 degrees Celsius
I have no clue, maybe someone else can help you
Answer: acceleration a = 25m/s^2
Explanation:
Given that:
The plane travels with constant acceleration
x1 = 241.22 m at t1 = 3.70 s
x2 = 297.60 m at t2 = 4.20 s
x3 = 360.23 m at t3 = 4.70 s.
We need to calculate the velocity in the two time intervals.
Interval 1:
Average Velocity v1 = ∆x/∆t = (x2 - x1)/(t2-t1)
v1 = (297.60-241.22)/(4.20-3.70) = 112.76m/s
Interval 2:
Average Velocity v2 = ∆x/∆t = (x3-x2)/(t3-t2)
v2 = (360.23-297.60)/(4.70-4.20)
v2 = 125.26m/s
Acceleration:
Acceleration a = ∆v/∆t
∆v = v2-v1 = 125.26m/s-112.76m/s = 12.5m/s
∆t = change in average time of the two intervals = (t3-t1)/2 = (4.70-3.70)/2 = 0.5s
a = 12.5/0.5 = 25m/s^2
