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Ber [7]
3 years ago
10

How can free runners extend the time of impact when they land on the ground after jumping long distance

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

By leaping.

Explanation:

When free runners run, they also leap. They do this so they can get more distance and air time. Leaping will allow them to get more distance per jump.

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ser-zykov [4K]

Answer:

Health-related physical fitness is primarily associated with disease prevention and functional health.

Explanation:

This is ur answer actually I just took it from Go ogle

7 0
3 years ago
Select all that apply so it’s not just one answer. please and thanks :))
qwelly [4]
Using sources that are peer reviewed
7 0
3 years ago
A new object, Object X, was discovered outside our solar system. Object X is small and rocky, is not a satellite of any other ob
hram777 [196]
<span>The characteristics of Object X most closely resemble those of other Comets. The correct option among all the options that are given in the question is the first option. The other options in the question can be negated. I hope that this is the answer that has actually come to your great help.</span>
4 0
3 years ago
Read 2 more answers
A 3-kg skateboard is rolling down the sidewalk at 4 m/s when it collides with a 1-kg skateboard that was initially at rest. If t
irga5000 [103]

Answer:

2 m/s

Explanation:

Parameters given:

Mass of first skateboard, m = 3 kg

Initial speed of first skateboard, u = 4 m/s

Mass of second skateboard, M = 1 kg

Initial speed of second skateboard, U = 0 m/s

Final speed of second skateboard, V = 6 m/s

Using the principle of the conservaton of momentum, the total initial momentum is equal to the total final momentum.

Momentum is the product of mass and velocity. This implies that:

m*u + M*U = m*v + M*V

(3*4) + (1*0) = (3*v) + (1*6)

12 + 0 = 3v + 6

=> 3v = 12 - 6

3v = 6

v = 6/3 = 2 m/s

The final speed of the 3 kg skateboard is 2 m/s

8 0
4 years ago
Read 2 more answers
The block accelerates down the 30 degree incline at 3m/s2. What is the coefficient of friction of these surfaces
dolphi86 [110]
Fnet = Fg sin 30 - Ff
ma = mg sin 30 - mew Fg cos 30
ma = mg sin 30 - mew mg cos 30
a = g sin 30 - mew gcos30
a - g sin 30 = - mew g cos 30
mew = -(a - g sin30)/(g cos 30)
mew = -(3m/s2 - 9.81sin30)/(9.81 cos 30)
mew = 0.22
8 0
3 years ago
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