1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find

2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:

The time taken by the pulse to travel from one support to the other is 0.208 s.
<h3>Given:</h3>
The mass of the cord is m = 0.65 kg.
The distance between the supports is, d = 8.0 m.
The tension in the cord is T = 120 N.
The time taken by the pulse to travel from one support to the other is given as,


Here, v is the linear velocity of a pulse. Its value is,



Then,


Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.
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Answer:
d) The 2 athletes reach the same height, because the athletes run with the same speed.
Explanation:
In the whole process , kinetic energy is converted into potential energy .
1/2 m v² = mgh
v² = 2gh
h = v² / 2g
In this expression we see that height attained does not depend upon mass of the object . At the same time it also makes it clear that it depends upon velocity . As the velocity in both the cases are same , height attained by both of them will be same. Hence option d ) is correct.
Answer:
Explanation:
If air resistance is ignored and assume UP and Toward Jason are the positive directions.
horizontal analysis
d = (vx₀)t
t = d/vx₀
horizontal analysis
0 = vy₀t + ½gt²
0 = vy₀(d/vx₀)+ ½g(d/vx₀)²
as vy₀ = v₀sin45 and vx₀ = v₀cos45 and are equal.
0 = d + ½g(d²/v₀²cos²45)
-d = ½g(d²/v₀²cos²45)
-dv₀² = ½g(d²/cos²45)
v₀² = -½g(d/cos²45)
v₀² = -½(-9.81(32.0/cos²45)
v₀² = 313.92
v₀ = 17.717787...
v₀ = 17.7 m/s