Answer:
A satellite on non-equatorial orbit would show daily motion even if its period is exactly 1 sidereal day.
Explanation:
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.
Try to have a calm morning before camp. A good night’s sleep and a good breakfast. Make sure to be cautious, follow all the rules for certain areas ( some maybe restricted ). Take lots of photos doing wacky stuff! Learn but have fun learning
Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m
Explanation:
the morning of the birthday party balloon filled with the 2.5 Litre of helium
temperature is 294kelvin
the party starts at the 4 p.m.
temperature rises 305 Kelvin.
the new volume = 4 litre.
At same temperature,
P
1
V
2
=P
2
V
2
(Boyle's law)
P
1
=10atm;P
2
=1atm
V
1
=4l=V
2
=8l
But while filling balloons from cylinder when pressure in cylinder becomes 1 atm then further filling is not possible (P
1
′
=9atm)
Let n be the number of balloons that can be filled.
∴P
1
′
V
1
=n(P
1
V
1
)
9×8=n(4×1)
n=
4
9×8
=18balloons