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abruzzese [7]
3 years ago
10

An object oscillates back and forth on the end of a spring. Which of the following statements are true at some time during the c

ourse of the motion?a. The object can have zero acceleration and, simultaneously, nonzero velocity.b. The object can have zero velocity and, simultaneously, zero acceleration.c. The object can have nonzero velocity and nonzero acceleration simultaneously.d. The object can have zero velocity and, simultaneously, nonzero acceleration.
Physics
1 answer:
Sindrei [870]3 years ago
8 0

Answer:

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

Explanation:

For an object in simple harmonic motion, the total mechanical energy (sum of elastic potential energy and kinetic energy) is constant:

E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2 (1)

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

We can also notice that the force on the spring is given by Hook's law:

F=-kx

And since according to Newton's law we have F = ma, this can be rewritten as

ma=-kx\\a=-\frac{k}{m}x

which means that the acceleration is proportional to the displacement.

So by looking again at eq.(1), we can now states that:

- when the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

- when the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

- in all the other intermediate situations, both velocity and acceleration are non-zero

So the correct answers are

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

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What force always attracts objects to each other ?
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A small artery has a length of 1.10 × 10-3 m and a radius of 2.50 × 10-5 m. If the pressure drop across the artery is 1.15 kPa,
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Answer:

7.69533\times 10^{-11}\ m^3/s

Explanation:

P = Pressure difference = 1.15 kPa

r = Radius = 2.5\times 10^{-5}\ m

\eta = Viscosity of liquid = 2.084\times 10^{-3}\ Pas

l = Length of artery = 1.1\times 10^{-3}\ m

From Poiseuille's equation we have

Q=\frac{\pi Pr^4}{8\eta l}\\\Rightarrow Q=\frac{\pi 1.15\times 10^3\times (2.5\times 10^{-5})^4}{8\times 2.084\times 10^{-3}\times 1.1\times 10^{-3}}\\\Rightarrow Q=7.69533\times 10^{-11}\ m^3/s

The flow rate of blood is 7.69533\times 10^{-11}\ m^3/s

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What do electricians call disconnects or disconnecting?
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1 year ago
A 57 kg boy and a 41 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio
Sergeeva-Olga [200]

Answer:

Acceleration of the boy a₁:

a_{1} = 1.87 \frac{m}{s^{2} }

Explanation:

Conceptual analysis

We apply Newton's second law to the boy and the girl:

F = m*a (Formula 1)

F : Force in Newtons (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Nomenclature

m₁ : boy mass

m₂ :  girl mass

a₁ : boy acceleration

a₂ :  girl acceleration

F₁ : boy acceleration

F₂ :  girl acceleration

Known data

m₁ =  57 kg

m₂ =  41  kg

a₂ = 2.6 m/s²

Problem development

We apply to Newton's third law of action and reaction, then:

F₁ = F₂ , We apply the formula (1):

m₁*a₁ = m₂*a₂

a_{1} = \frac{m_{2}* a_{2} }{m_{1} }

a_{1} = \frac{41* 2.6 }{ 57 }

a_{1} = 1.87 \frac{m}{s^{2} }

5 0
3 years ago
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