A muscle that assists the prime mover or helps stabilize the joint during movement would be ________

1 answer:

5 0

A muscle that assists the prime mover or helps stabilize the joint during movement would be a synergist.

Hope I helped!

You might be interested in

1. ¿Qué presión se ejerce sobre cada una de las cuatro patas de una mesa si su masa es de 20 kg y

mestny [16]

Tenemos.

Masa de la mesa = 20kg
Masa encima = 10kg
Masa total = 30kg
Area de cada pata = 20cm² = 20cm² * 1/10000cm² * 1m² =0,002m²

Presió(P)n que se ejerce sobre cada pata.

P =F/A
P = Masa por gravedad/A Masa = 30kg Gravedad =9,8m/s²
P =(30kg * 9,8m/s²)/0,002m²
P = 294kg * m/s²/0,002m² Pero kg *m/s² = Nw
P = 294Nw0,002m²
P = 147000 Nw/m² Pero Nw/m² = pa
p =1,47 * 10⁵ pa

Respuesta.
La presión que se ejerce sobre cada pata es de 1,47 * 10⁵pa

8 0

2 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$