Which excerpt are you talking about?
Answer:
<em>2 m/s</em>
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Explanation:
The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

where
is the induced voltage = 2.88 mV = 2.88 x 10^-3 V
is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m
is the velocity of the fluid through the field = ?
is the magnetic field = 0.120 T
substituting, we have
2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x 
2.88 x 10^-3 = 1.44 x 10^-3 x 
= 2.88/1.44 = <em>2 m/s</em>
Bro the md that she lost was the md boneless burger
Answer:
B and C!
Explanation:
I just did it on edge. I hope this helps!
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).