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Pie
3 years ago
12

Match the major muscle group with its functional role.

Physics
1 answer:
Naily [24]3 years ago
7 0

Answer:

Abdominal

Sitting up, postural alignment

Biceps

Lifting, pulling

Deltoids

Overhead lifting

Erector Spinae

Postural alignment

Gastronemius & Soleus

Push off for walking, standing on tiptoes

Gluteus

Climbing stairs, walking, standing up

Hamstrings

Walking

Latissimus Dorsi & Rhomboids

Postural alignment, pulling open a door

Obliques

Rotation and side flexion of body

Pectoralis

Push up, pull up, bench press

Quadriceps

Climbing stairs, walking, standing up

Trapezius

Moves head sideways

Triceps

Pushing

God bless you. Because my soul almost left my body when i had to do this.

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An elevator has a mass of 1000 Kg. What force is needed to accelerate it upward at a rate of 2 m/s/s?
zubka84 [21]

The force needed to accelerate an elevator upward at a rate of 2 m / s^{2} is 2000 N or 2 kN.

<u>Explanation: </u>

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as 2 m / s^{2}, the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

                  \text {Force}=\text {Mass} \times \text {Acceleration}

                  \text { Force }=1000 \times 2=2000 \mathrm{N}=2 \text { kilo Newon }

So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of 2 m / s^{2}.

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3 years ago
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

   Hence

        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

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