a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
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Answer:
F = 878.9 N
Explanation:
The electrostatic force of attraction or repulsion is given by Coulomb's Law as follows:
F = kq₁q₂/r²
where,
F = Force pf repulsion between balloons = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = q₂ = magnitudes of 1st and 2nd charge = 0.0025 C
r = distance between balloons = 8 m
Therefore,
F = (9 x 10⁹ N.m²/C²)(0.0025 C)(0.0025 C)/(8 m)²
<u>F = 878.9 N</u>
B) using a blood pressure monitor to find a patient's blood pressure.
Hope this helped!
