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Natalija [7]
3 years ago
10

Which of the following is true about a rigid body in dynamic equilibrium? The body can have translational motion, but it cannot

have rotational motion. The body can have translational motion and rotational motion, as long as its translational and angular accelerations are equal to zero. The body cannot have translational motion, but it can have rotational motion. The body cannot have translational or rotational motion of any kind.
Physics
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

The correct answer is "The rigid body can have rotational and transnational motion, as long as it's transnational and angular accelerations are equal to zero."

Explanation:

A rigid body by definition does not deform when forces act on it. In case of static equilibrium a rigid body cannot have any sort of motion while in case of dynamic equilibrium it can move but with constant velocities only thus having no acceleration weather transnational or angular.

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If a an object is traveling at a rate of 5 meters per second squared with a force of 10 Newtons, what is the mass of the object?
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Answer:

2 kg

Explanation:

Acceleration = 5 m/s^2

Force = 10 N

Force = mass * acceleration

mass = force / acceleration

mass = 10 / 5

mass = 2 kg

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A motorbike accelerates from 15m/s to 25m/s in 15 seconds.
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distance traveled by a uniformly accelerated bike is given as

d = \frac{v_f + v_i}{2} (t)

here we know that

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v_i = 15 m/s

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Compare and contrast model A with model C. How are they alike: How are they different? A) They are composed of different element
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The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within
zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, m=9.1\times 10^{-31}\ kg

Initial speed of electron, u = 0

Final speed of electrons, v=3\times 10^7\ m/s

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2}{2\times 0.063}

a=7.14\times 10^{15}\ m/s^2

Now we will find the electric field required in the tube as :

ma=qE

E=\dfrac{ma}{q}

E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

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2 years ago
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