Answer:I need the ans to this one too
Explanation:
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:
m= 10 kg a = 52 m / s²
Explanation:
For this problem we must use Newton's second law, let's apply it to each axis
X axis
F - fr = ma
The equation for the force of friction is
-fr = miu N
Axis y
N- W = 0
N = mg
Let's replace and calculate laceration
F - miu (mg) = ma
a = F / m - mi g
a = 527.018 / m - 0.17 9.8
We must know the mass of the body suppose m = 10 kg
a = 527.018 / 10 - 1,666
a = 52 m / s²
