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lana [24]
2 years ago
7

What is an essential characteristic of an object in equilibrium?

Physics
1 answer:
bulgar [2K]2 years ago
5 0

Answer:

2) zero acceleration

Explanation:

Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

Mathematically, the motion of an object is described in terms of time, distance, speed, velocity, position, displacement, acceleration, etc.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

Generally, an object is said to be in equilibrium when neither the energy possessed by the object not state of motion changes with respect to time. Thus, the vector sum of all the forces acting upon an object that's in equilibrium is zero.

In conclusion, an essential characteristic of an object in equilibrium is zero (0) acceleration because there's no change in its velocity with respect to time.

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How many centimeters are in 3.50 feet?
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106.68 centimetres are in 3.50 feet
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Pllssss answer thiss ...<br> its about the reading of VOLTMETER AND AMMETER
Afina-wow [57]

Answer:

The answer is C 1.8V and 0.38A

4 0
2 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
2 years ago
A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg
motikmotik

Answer:

\omega_{f}=1.634\ rad/s  

Explanation:

given,  

diameter of merry - go - round = 2.40 m  

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg  

initial angular momentum of the system  

L_i = I\omega_i  

L_i =356\times 1.80  

L_i =640.8\ kg.m^2/s  

final angular momentum of the system  

L_f = (I_{disk}+mR^2)\omega_{f}  

L_f = (356 + 25\times 1.2^2)\omega_{f}  

L_f= (392)\omega_{f}  

from conservation of angular momentum  

L_i = L_f  

640.8= (392)\omega_{f}  

\omega_{f}=1.634\ rad/s  

8 0
2 years ago
Read 2 more answers
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