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3 years ago

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If the net external torque acting on the particle is zero, the it satisfies - 1st condition of equilibrium - 2nd condition of eq

uilibrium - Both 1st & 2nd condition of equilibrium - 2nd law of motion

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

a) the distances are zero, Both 1st & 2nd condition

c) the torques are equal but of the opposite sign, 2nd condition of equilibrium

Explanation:

The equilibrium conditions are

1 translational

∑ F = 0

2 rotational

∑ τ = Σ (F_i x r_i) = 0

They tell us that external torque is zero.

Therefore we have two various possibilities

a) the distances are zero, in this case we have a pure translation movement

for this situation the two equilibrium relations are fulfilled

b) the forces are zero, there is no movement

It does not make sense to use the equilibrium relations since there are no forces

c) the torques are equal but of the opposite sign, the forces are on the opposite side of the body.

In this case the 2 equilibrium relation is fulfilled, but not the first one that the force has the same direction

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The binding of acetylcholine to its receptor at the neuromuscular junction causes the opening of a

Rufina [12.5K]

Answer: opening of the nicotinic acetylcholine receptor channels.

Explanation:

Neuromuscular junction is a special junction formed between a motor neurone and a muscle fibre. The junction is fortified with nerves and receptors that helps in the transmission of signals from the motor neurone to the muscle fibre in order to bring about the desired voluntary movements through muscular contraction.

Nicotinic acetylcholine receptor are activated through the binding of acetylcholine at the neuromuscular junction. This action leads to influx of sodium ions to accomplish endplate potential.

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3 years ago

A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

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3 years ago

which best describes what happens with the sonar to sound pulses from a ship after they hit the ocean floor?

lakkis [162]

Answer:

With sonar, what happens to sound pulses from a ship after they hit the ocean floor? ... They bounce back to the ship.

Explanation:

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3 years ago

In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for

4vir4ik [10]

Answer:

The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is

$k = 0.903$

Explanation:

From the question we are told that

The time constant $\tau = 3$

The potential across the capacitor can be mathematically represented as

$V = V_o (1 - e^{- \tau})$

Where $V_o$ is the voltage of the capacitor when it is fully charged

So at $\tau = 3$

$V = V_o (1 - e^{- 3})$

$V = 0.950213 V_o$

Generally energy stored in a capacitor is mathematically represented as

$E = \frac{1}{2 } * C * V ^2$

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now since capacitance is constant at $\tau = 3$

The energy stored can be evaluated at as

$V^2 = (0.950213 V_o )^2$

$V^2 = 0.903 V_o ^2$

Hence the fraction of the energy stored in an initially uncharged capacitor is

$k = 0.903$

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3 years ago

How is light refracted inside a crystal ball ?

Bingel [31]

The light refects ofthe sides

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3 years ago