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Serjik [45]
2 years ago
8

To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular accelera

tion. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant, nonzero angular acceleration. The kinematic equations for such motion can be written as
Physics
1 answer:
Elis [28]2 years ago
5 0

Answer:

   ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

Explanation:

Rotational kinematics can be treated as equivalent to linear kinematics, for this change the displacement will change to the angular displacement, the velocity to the angular velocity and the acceleration to the angular relation, that is

     x → θ

     v → ω

     a → α

with these changes the three linear kinematics relations change to

      ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

where it should be clarified that to use these equations the angles must be measured in radians

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Explanation:

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8 0
3 years ago
if a cat is running at a constant speed of 10km/h for 5 s, what is its average speed and what is its instantaneous speed at 4 s?
iris [78.8K]

Here a cat is running at constant speed which is given as 10 km/h for  5s

So here the average speed is defined as total distance moved in total time interval

so here it is given by

v_{avg} = \frac{distance}{time}

since

here speed of cat is constant so it will remain the same

And hence the average speed and instantaneous speed at any instant for this duration will remain the same

so here answer would be

<em>average speed = 10 km/h</em>

<em>instantaneous speed = 10 km/h</em>

8 0
3 years ago
A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f
uysha [10]

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

\Sigma F=ma

Replacing the values:

mg\sin 40=ma

We may cancel out the mass:

g\sin 40=a

Using the gravity constant as 9.8 meters per square second:

9.8\frac{m}{s^2}\sin 40=a

Solving the operations:

6.3\frac{m}{s^2}=a

Therefore, the acceleration is 6.3 meters per square second.

8 0
1 year ago
A coil formed by wrapping 65 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the
Fiesta28 [93]

Answer:

377 m

Explanation:

number of turns, N = 65

θ = 36°

B1 = 200 micro Tesla

B2 = 600 micro tesla

t = 0.4 s

induced emf, e = 80 mV

Let a be the side of the square coil.

e=\frac{d\phi }{dt}=NA\frac{dB}{dt}\times Sinθ

0.080=\frac{65\times a^{2}\times Sin36\times\left ( 600 - 200 \right )\times 10^{-6}}{0.4}

0.080=0.038a^{2}

a = 1.45 m

Total length of the wire, L = N x 4a = 65 x 4 x 1.45 = 377 m

Thus, the length of the wire is 377 m.

7 0
3 years ago
A tortoise can run with a speed of 0.14 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, b
Reptile [31]

speed of tortoise is given as v1 = 0.14 m/s

speed of hare is given as v2 = 20*0.14 = 2.8 m/s

now let say the total length of the path is "d"

so the total time taken by the tortoise to cover this

t = \frac{d}{0.14}

now given that hare took rest for 1 min

so total time of run for hare is (t - 60)s

so the distance that hare covered is given by

d - 0.30 = 2.8 * (t - 60)

now by above two equations

d - 0.30 = 2.8 * \frac{d}{0.14} - 168

168 - 0.30 = (20 - 1)d

d = 8.82 m

and the time t is given by

t = \frac{8.82}{0.14}

t = 63 s

so part a)

t = 63 s

part b)

d = 8.82 m

4 0
3 years ago
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