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2 years ago

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

2 answers:

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]2 years ago

4 0

This is because it is the prime mover in both cases. 2. Antagonist: The antagonistin a movement refers to the muscles that oppose the agonist. During elbow flexionwhere the bicep is the agonist, the tricep muscle is the antagonist.

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The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

3 years ago

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

6 0

3 years ago

A charged balloon will stick to a neutral wall. which process is involved?

Leokris [45]

The charged balloon will stick to a neutral wall because of the Static Electricity:

The matter is formed by atoms and these atoms are composed of electrons, protons and neutrons (the electrons have a negative charge, the protons have a positive charge and the neutrons have no charge).

As the balloon is charged (It gained electrons), and the charge of the same sign repel each other, when it approaches the wall, the electrons of this wall will move away, and the positive charges (protons) will remain in the nearest area to the balloon. As the charges of different signs are attracted, the balloon will be stuck to the wall.

3 0

3 years ago

If something is traveling at 20 m/s constant velocity, is it in equilibrium? If a projectile is launched at a velocity of 20 m/s

Irina18 [472]

If something is traveling at 20 m/s constant speed AND its direction isn't changing, then its velocity is constant. Another way to say that is: Its acceleration is zero. Zero acceleration means zero NET force acting on the object, or a group of BALANCED forces acting on it, also called EQUILIBRIUM. The required answer is: YES.

If a real projectile is launched, the force of gravity acts on it vertically downward. There's no upward force acting on it to balance gravity. Therefore, the forces on the projectile are NOT balanced, there IS a net vertical force on it, and it's NOT in equilibrium. Too bad.

3 0

2 years ago

The bob of a pendulum swings back and forth with a total mechanical energy of 300 J. What is the kinetic energy of the bob when

zhenek [66]

at the lowest point in the trajector, the kinetic energy of the bob is 300 J.

Explanation:

The total mechanical energy of the bob at any point of its motion is given by

$E=KE+PE$

Where

$KE=\frac{1}{2}mv^2$ is the kinetic energy, where

m is the mass of the bob

v is its speed

$PE=mgh$ is the gravitational potential energy, where

g is the acceleration of gravity

h is the height of the bob, measured with respect to the lowest point of the trajector

In absence of friction, the total mechanical energy E remains constant. So we have:

- When the bob swings upward, the PE increases (because h increases) and the KE decreases (so the speed decreases). At the highest point in the trajector, the speed of the bob is zero (v=0), so its KE is also zero and all the mechanical energy is potential energy: U = 300 J

- When the bob swings downward, the PE decreases (because h decreases) and the KE increases (so the speed increases). At the lowest point in the trajectory, the height has become zero (h=0), so the PE is zero and all the mechanical energy is kinetic energy: KE = 300 J

Therefore, at the lowest point in the trajector, the kinetic energy of the bob is 300 J.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

4 0

3 years ago