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IrinaK [193]
3 years ago
15

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

your wrist as you grip it between your thumb and forefinger with a force of 5.0 N . Suppose the coefficient of static friction between the knob and your fingers is 0.65.
What is the most torque that you can exert on the knob without having it slip between your fingers?
Physics
1 answer:
galina1969 [7]3 years ago
4 0

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

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When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi
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Answer:

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Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este
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a) Density of the liquid: 823.5kg/m^3

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

<em>"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate: </em>

<em>a) the density of the liquid; </em>

<em>b) the weight of the liquid."</em>

a)

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

M=m_L + m_V (1)

where

m_L is the mass of the liquid

m_V is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

m_L=M-m_V=175-35=140 kg

The density of the liquid is given by

d=\frac{m}{V}

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = 0.170 m^3 (volume of the liquid, which is equal to the volume of the vessel)

So we get

d=\frac{140}{0.170}=823.5kg/m^3

b)

The weight of a body is given by

F=mg

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

g=9.8 m/s^2 (acceleration due to gravity)

Therefore, its weight is

F=(140)(9.8)=1372 N

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shepuryov [24]

1.

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force of earth's gravity on Mr. Ure is given as

F = mg

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2.

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g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

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F = mg

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