Russ makes the diagram below to organize his notes about how Newton’s first law describes objects at equilibrium.

Physics

2 answers:

viva [34]2 years ago

8 0

Correct options:

Has a net force of zero

Explanation:

Newton's first law, also known as law of inertia, states that:

"An object at rest which has a net force of zero acting on it stays at rest, and an object moving at constant velocity which has a net force of zero acting on it continues its motion at constant velocity"

The first situation (object at rest) is called static equilibrium, while the second situation (object moving at constant velocity) is called dynamic equilibrium. As we read from the statement of the Newton's first law, in both situations the object has a net force of zero acting on it, so this is the label that should be placed in region X (which is the region in common between both dynamic and static equilibrium)

tiny-mole [99]2 years ago

6 0

Newton's first law states that an object at rest remains so until acted upon by external forces or an object in motion does so without accelerating (that is, at constant speed).

Therefore, x can only describe an object coming to rest. That, the last option is more correct.

You might be interested in

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$