opted solution

[116903]

---

Members do not see this ad.

😕Is there any way i can find a solution for opted questions?
I would like to get explanation for physics part?
thank you oin advance.

 

[116903]

can anyone solve below two opted questions?

12. a satellite of mass 200 kg completes a circular orbit of the earth in 120mins. a 400 kg satellite is then put into orbit at the same height above the earth. which of the following represents the time, in minutes, for the 400 kg satellite to complete a circular orbit?

16. a ball is horizontally thrown at 20m/s from the top of a building and strikes the level ground 50 m from the building. how many meters high is the roof of the building?

 

[Commando303]

12. I forgot.

16. I think the building is 30.625 meters high.

 

[EyEnStein 07]

can anyone solve below two opted questions?

16. a ball is horizontally thrown at 20m/s from the top of a building and strikes the level ground 50 m from the building. how many meters high is the roof of the building?

I think you would care for the explanation to 2...so here it is.

+a ball is thrown horizontally 20m/s...this means that V0x = 20m/s
+from the top of a building...if we draw a diagram...label the building "h" for height.
+it strikes ground 50m away from the building...so that means if we take our coordinate system origin at the building (0,0)..then the change X-X0 = 50.

To find how high the building is...we have to look back to one of the kinematic equations. We should first find out the time it took to strike the ground.

1) x-x0 = Vox * t
50m = 20m/s * t --> the "m" cancels...we have [5/2 s = t]

Now that we have the time, we can use the kinematic equation that allows us to find height H, in terms of y-y0
2) y-y0 = V0y*t + 1/2*g*t^2
we know that y-y0 is going to be the height...so we can replace that with "h"

h=v0y*t + 1/2*g*t^2
v0y = 0....the ball was released horizontally only...which means it only has a horizontal component, v0y is a vertical component...you can also see this because gravity takes over after its initial 20m/s horizontal.

h= 0 + 4.9m/s^2 (5/2s)^2 --> the seconds cancel out
h=30.6 m

 

[sugarhype10]

Maybe Kepler's Third Law equation for satellite orbit period will be helpful. It is:

T= sq.rt[(4pi^2R^3)/(GM)]

R= radius of orbit for satellite

G is the universal gravitational constant
G = 6.6726 x 10-11N-m2/kg2

M = mass of the central body

 

[NeverGiveUp]

I think you folks should seriously use OAT Achiever, which greatly helped with my second OAT scores. It's the only CBT with detailed explanations, and has some questions you can look at through the DEMO here:

http://www.3tquest.com/Vista/Instruction.htm

 

[sugarhype10]

I just took the opted exam myself, and here's my reasoning for #12 of the physics section:

Looking at Kepler's third law, you see that the time it takes to complete a circular orbit depends on the gravitational constant, the mass of the central body, and the radius. The time it takes to complete a circular orbit does not depend on the mass of the satellite. Since the height is the same, the time is still 120 days.

 

[NatashaColorado]

I think you would care for the explanation to 2...so here it is.

+a ball is thrown horizontally 20m/s...this means that V0x = 20m/s
+from the top of a building...if we draw a diagram...label the building "h" for height.
+it strikes ground 50m away from the building...so that means if we take our coordinate system origin at the building (0,0)..then the change X-X0 = 50.

To find how high the building is...we have to look back to one of the kinematic equations. We should first find out the time it took to strike the ground.

1) x-x0 = Vox * t
50m = 20m/s * t --> the "m" cancels...we have [5/2 s = t]

Now that we have the time, we can use the kinematic equation that allows us to find height H, in terms of y-y0
2) y-y0 = V0y*t + 1/2*g*t^2
we know that y-y0 is going to be the height...so we can replace that with "h"

h=v0y*t + 1/2*g*t^2
v0y = 0....the ball was released horizontally only...which means it only has a horizontal component, v0y is a vertical component...you can also see this because gravity takes over after its initial 20m/s horizontal.

h= 0 + 4.9m/s^2 (5/2s)^2 --> the seconds cancel out
h=30.6 m

See, I would have rounded 9.8 gravity to 10 (like I was advised, for faster short hand calculations), and my answer would be 31.25m. Hopefully, the test answers are far enough apart.

 

[optmania]

Can anyone help me with these physics problems on the OPTED


25. A uniform block of mass 180g that is 10 X 9 X 3 cm is to be placed in a liquid of density .900g/cm^3. the block will

a. sink in the liquid
b. just float in the liquid above with none of its volume aboce the surface of the liquid
c. float in the liquid with more that 1/2 of its volume above the surface of the liquid
d. float in the liquid with less that 1/3 of its volume above the surface of the liquid
e. float in the liquid with all of its volume above the surface of the liquid

21.the amplitude of a body undergoing simple harmonic motion is doubled. which of the following is also doubled

a.maximum speed
b.frequency
c.mass
d.total energy
e.period

13. Atmospherid pressure is very nearly 100kPA. A sealed container of air at 1 atmospheric pressure has a door 1m wide and 2m high. This door is very hard to open during high pressure days. if the atmospheric pressure on the outside of the container is just 1 percent greater than on the inside then what added force in newtons is rewquired to open the container door
a. 100
b. 200
c. 2.000
d.10000
e. 20000

As you guys can see im not the greatest in the fluid problems. Hope someone can help me