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givi [52]
3 years ago
7

What are the directions of movement for the Sagittarius, frontal, transverse planes ?

Physics
1 answer:
Dima020 [189]3 years ago
5 0

Answer:

Squats involve flexion (forward motion) and extension (backward on the way up), so would fit into the sagittal plane. Frontal plane motion would include leaning from left to right as in sidebends and lateral raises, or perhaps you might picture jumping jacks for a good image of movement along the frontal plane.

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What was the first dogs name that landed on the moon
elena55 [62]

Answer:

Laika

Explanation:

The dog's name was originally Kudryavka, or Little Curly, but she became known internationally as Laika, a Russian word for several breeds of dog similar to a husky. American reporters dubbed her Muttnik as a pun on Sputnik. Unfortunately, Laika's trip into space was one-way only.

8 0
3 years ago
What is the name of the system for conducting a scientific investigation?
Slav-nsk [51]

Answer:

i beleive it is 2

Explanation:

5 0
3 years ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
3 years ago
A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

     y =v_{oy} t -1/2 g t²

     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

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     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

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     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0
4 years ago
2 When a cube of hot metal is placed in a beaker of cold water, the temperature of the water -
jek_recluse [69]
The answer to this is B
4 0
3 years ago
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