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geniusboy [140]

3 years ago

12

A particle moving along the x-axis has a position given by x = (24t – 2.0t 3 ) m, where t is measured in s. What is the magnitud

e (m/s2 ) of the acceleration of the particle when the particle is not moving?

1 answer:

Vitek1552 [10]3 years ago

3 0

<h2>The magnitude 24 ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.</h2>

Explanation:

Given,

A particle moving along the x-axis has a position given by

$x=(24t-2.0t^3)$ m ........ (1)

To find, the magnitude ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving = ?

Differentiating equation (1) w.r.t, 't', we get

$\dfrac{dx}{dt} =\dfrac{d((24t-2.0t^3))}{dt}$

⇒ $\dfrac{dx}{dt} =24(1)-3(2.0)t^{2} =24-6t^{2}$ ....... (2)

⇒ $24-6t^{2} = 0$

⇒ $t^{2}=2^{2}$

⇒ t = 2 s

Again, differentiating equation (2) w.r.t, 't', we get

$\dfrac{d^2x}{dt^2} =-12t$

Put t = 2, we get

$\dfrac{d^2x}{dt^2} =-12(2)=24$

Thus, the magnitude 24 ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.

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Describe the astronomical achievement of Mesopotamia

Annette [7]

Beginning around 5,500 years ago, the Sumerians built cities along the rivers in Lower Mesopotamia, specialized, cooperated, and made many advances in technology. The wheel, plow, and writing (a system which we call cuneiform) are examples of their achievements.

3 0

2 years ago

A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.

Anestetic [448]

Answer:

$\omega=0.37 [rad/s]$

Explanation:

We can use the conservation of the angular momentum.

$L=mvR$

$I\omega=mvR$

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

$(I_{proffesor - stool}+mR^{2})\omega=mvR$

Now, we just need to solve it for ω.

$\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}$

$\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}$

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I hope it helps you!

5 0

3 years ago

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed

beks73 [17]

Answer:

v = 5.9 x 10⁷ m/s

Explanation:

The kinetic energy of the electron in terms of potential difference is given as:

$K.E = eV$ --------------- equation (1)

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

V = Potential Difference = 9.9 KV = 9900 Volts

The kinetic energy in general is given as:

$K.E = \frac{1}{2}mv^{2}\\$ --------- equation (2)

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

Therefore, comparing equation (1) and equation (2), we get:

$\\\frac{1}{2}mv^{2} = eV\\\\\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)v^{2} = (1.6\ x\ 10^{-19}\ C)(9900\ volts)\\\\v = \sqrt{34.81\ x\ 10^{14}} \\$

<u>v = 5.9 x 10⁷ m/s</u>

8 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

2 years ago

1. Calcula la fuerza de atracción electrostática entre dos cuerpos de cargas q1 = -18 C y q2 = +5 mC, separados entre sí por una

romanna [79]

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

q2=5mC=0.005C

d=25cm=0.25m

Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.

Thus:

F=9×10⁹×(-28)×0.005/0.25²

F=-20.16×10⁹N

The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

6 0

3 years ago