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Hunter-Best [27]
2 years ago
15

Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to

be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero
Physics
1 answer:
jok3333 [9.3K]2 years ago
7 0

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

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

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aleksklad [387]

Answer:

35.7 m

Explanation:

Let

\mid A\mid=18.5 m

\mid B\mid=41 m

We have to find the distance between Joe's and Karl'e tent.

A_x=Acos\theta

A_y=Asin\theta

Substitute the values then we get

A_x=18.5cos23^{\circ}=17 m

A_y=18.5sin 23^{\circ}=7.2 m

B_x=41cos37.5^{\circ}=32.5 m

B_y=41sin37.5^{\circ}=-24.96 m

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

B=A+C

C=B-A

C_x=B_x-A_x=32.5-17=15.5 m

C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m

\mid C\mid=\sqrt{C^2_x+C^2_y}

\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m

Hence, the distance between Joe's and Karl's tent=35.7 m

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The only 'difference' is that they are different categories.

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