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antiseptic1488 [7]
3 years ago
10

At which moon position would a person on Earth see the entire half of the moon(full-moon)?

Physics
2 answers:
iragen [17]3 years ago
8 0
Wouldn't that be the full moon position...its position 5...idk know what its asking for exactly..what are your choices?

jeka943 years ago
3 0
Sun-earth-moon in a straight line. Earth in the 'middle'.
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Determine in symbols an expression for the magnetic force exerted on the falling bar (and determine the direction of that force)
Katarina [22]

Answer:

magnetic force on falling Bar F = B*i*L*sin(90) = B*(B*L*v/R)*L = B^2*L^2*v/R

direction of the force is vertically upwards

Explanation:

8 0
4 years ago
How does the sun's gravity and the earth inertia keep us orbiting in the solar system
scZoUnD [109]
<span>Inertia keeps us orbiting because any object with mass has the tendency to resist changes to their direction and speed of movement. Combine that with the interaction of the gravitational attraction of the sun, and that is what keeps Earth in orbit. The sun’s gravitational force is one that is proportional to Earth’s mass, and it acts in a way that is almost exactly perpendicular to Earth’s motion. This keeps Earth from spinning into the sun or far away from it.</span>
6 0
4 years ago
Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

5 0
3 years ago
Donde esta la orilla del universo?
irinina [24]

Answer:

<em>El universo no tiene un borde finito</em>, continúa y se expande sin fin. La distancia adecuada desde la Tierra hasta el borde del universo observable es de aproximadamente 46.5 mil millones de años luz o 4.40 × 10 ^ 26 metros en cualquier dirección.

5 0
3 years ago
The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of
PolarNik [594]
The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions
7 0
3 years ago
Read 2 more answers
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