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Mashutka [201]
3 years ago
10

What are the agonist and antagonist muscle of a wrist joint???

Physics
2 answers:
oksian1 [2.3K]3 years ago
4 0
Wrist flexors are the agonist muscles, while wrist extensors are the muscle antagonists. The specific names are the flexor digitorum and the extensor digitorum.
Andreas93 [3]3 years ago
4 0
This is because it is the prime mover in both cases. 2. Antagonist<span>: The </span>antagonist<span>in a movement refers to the muscles that oppose the agonist. During </span>elbow flexion<span>where the bicep is the agonist, the tricep muscle is the </span>antagonist<span>.</span>
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Find the volume of a box measuring 2 cm by 7 cm by 3 cm.
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What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight ou
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Five properties of magnet​
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3 years ago
An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
oksian1 [2.3K]

Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

b)

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

d)

Spring Constant:

As w=\sqrt{k/m}

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

⇒ v= 8.78 m/s

f)

Maximum force will be exerted on the block when it is at maximum distance.

F= k* A   ( A is amplitude of oscillation)

F= 179.2 * 0.27 N

F= 48.4 N

5 0
3 years ago
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