The air inside becomes warmer, and that makes it LESS DENSE.
Less-dense air FLOATS in denser air. Since it's inside the balloon, it picks the balloon up when it floats.
Answer:
3.626 m/s
Explanation:
v=d/t
1. -0.02/0 = 0 m/s
2. 0.86/0.2 = 4.3 m/s
3. 1.71/0.4 = 4.275 m/s
4. 2.54/0.6 = 4.23 m/s
5. 3.32/0.8 = 4.15 m/s
6. 4.08/1.0 = 4.08 m/s
7. 4.79/1.2 = 3.99 m/s
8. 5.48/1.4 = 3.91 m/s
9. 6.15/1.6 = 3.84 m/s
10. 6.76/1.8 = 3.76 m/s
11. 7.37/2.0 = 3.66 m/s
12. 7.92/2.2 = 3.6 m/s
13. 8.45/2.4 = 3.52 m/s
14. 8.96/2.6 = 3.45 m/s
the mean of these numbers is 3.626
his average velocity ks 3.626 m/s
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:
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The driver is tooling along in his snowmobile, pointed north,
at 8.5 m/s.
He's carrying the flares with him, so the flares are also moving north
at 8.5 m/s.
When he fires the flare straight up, it has a vertical velocity of 4.3 m/s
straight up, and a horizontal velocity of 8.5 m/s towards the north.
The magnitude of the net velocity is √(4.3² + 8.5²) .
That's about 9.53 m/s, at some angle between straight up
and straight north.
The angle above horizontal is the angle that has a tangent of 4.3/8.5 .
I'll let you work out the angle.
False because faster moving slope would be going up and slower would be going down because its decreasing
