What is first and second conditions of equilibrium?

A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist

galina1969 [7]

Answer:

0.04225 Nm

Explanation:

N = Force applied = 5 N

$\mu$ = Coefficient of static friction = 0.65

d = Diameter of knob = 1.3 cm

r = Radius of knob = $\frac{d}{2}=\frac{1.3}{2}=0.65\ cm$

Force is given by

$F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N$

When we multiply force and radius we get torque

Torque on thumb

$\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm$

Torque on forefinger

$\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm$

The total torque is given by

$\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm$

The most torque that exerted on the knob is 0.04225 Nm

4 0

2 years ago

Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw

Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

$\displaystyle{U=\frac{q}{4\pi \epsilon_0r}}$ (1)

where $q$ is the charge of the particle, $\epsilon_0$ the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and $r$ is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

$\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}$

Substituting the values $q=1.602 \cdot10^{-19}\ C$ , $\displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}}$ and $r=0.5 \cdot 10^{-9} m$ we obtain:

$\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}$

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0

3 years ago

A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same

NISA [10]

Answer:

The box arrives first.

Explanation:

Hope this helps!! :))

Learn more at :

brainly.com/question/20164917

4 0

2 years ago