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Katen [24]
3 years ago
15

Give an example from sport where at least 3 fundamental skills are linked together, name the movement action and name the three

Fundamental movement skills
Physics
1 answer:
Anika [276]3 years ago
8 0
Basketball,
Dribbling, running, shooting
Running, passing, jogging
You might be interested in
A sodium has an atomic number of 11 and the atomic mass of 23. Determine the its proton, electrons, and neutrons
ioda

The mass number of an element tells us the number of protons AND neutrons in an atom (the two particles that have a measurable mass). Sodium has a mass number of 23amu. Since sodium has 11 protons, the number of neutrons must be 23 – 11 = 12 neutrons.

7 0
3 years ago
A woman takes her dog Rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho
ValentinkaMS [17]

Answer:

The force parallel to the horizontal is 26.24 N

Explanation:

She pulls on the leash with a force F = 30 N, this force, since its at an angle of 29° (i will cal this angle \theta), it has a force component on x (the horizontal, i will call this force F_{x}) and a force component on y (the vertical, i will call this F_{y} ).

This can be seen in the attached picture.

Since we are asked about the force parallel to the horizontal, we need to find the component of the force F_{x}, since F_{x} is the adjacent angle, we need to use cosine:

F_{x}=Fcos \theta

since F=30N and \theta=29

F_{x}=(30N)cos(29)

F_{x}=(30N)(0.8746)

F_{x}=26.24N

The force parallel to the horizontal is 26.24 N

7 0
3 years ago
4. In which of the following circuits will the bulb glow?
Zanzabum

Answer:

I believe is C.

Explanation:

its quite obvious:)

4 0
3 years ago
PLEASE HELP ME
ra1l [238]
C should be the answer
8 0
3 years ago
Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?
Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

x_{a}(t) = at+bt^2

x_{b}(t) = ct^2-dt^3

Constants are:

a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both x_a(t) and x_b(t) depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

at+bt^2 = ct^2-dt^3

2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0

t_1 = 4 - \sqrt{3} = 2.27 s, t_1 = 4 + \sqrt{3} = 5.73 s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0

t_1 = 1 s, t_2 = 4.33 s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67 s

3 0
4 years ago
Read 2 more answers
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