Answer:

Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So
and
(Since
). Then, we get:

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).
Answer:
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
Explanation:
As we know that the relation between electric field and electric potential is given as

here if we say that potential is constant because electric field sensor is moving along equi-potential line.
Then we will say
V = constant
so we have

so electric field will remain constant always in magnitude and always remains perpendicular to the surface
so we have
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
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<span>a=Δω/Δt
</span><span>a=2π*Δf/Δt
</span><span>a=2π*(f2-f1)/Δt
</span>
<span>f1=f2-a*Δt/2π
</span><span>f2=800/60 rev/sec
</span><span>a=-42 rad/sec^2
</span><span>Δt=1.75sec
</span><span>so
f1=25 rev/sec
f1=1500 rev/min</span>
Answer:
= 9.8°
Explanation:
Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.
width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m
angular position of fringe, Sinθ = n λ /a
n is order of fringe , λ is wave length of light and a is slit aperture
So Sinθ ∝ 1 / a
Sin θ₁ /Sin θ₂ = a₂/a₁ ;
Sin20°/sinθ₂ = 2 / 1
sinθ₂ = Sin 20° / 2 = .342/2 = .171
θ₂ = 9.8 °
Answer:
the answer from my side is both vertical and horizontal