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Ugo [173]
3 years ago
5

Is magnitude a vector scalars both neither

Physics
1 answer:
Doss [256]3 years ago
7 0

Answer:

scalars

Explanation:

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A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
where \omega_i and \omega_f are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
\alpha =  \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
where \omega_i = 0 is the initial angular speed. So, the angle covered after a time t=3.07 s is
\theta=  \frac{1}{2}  \alpha t^2 =  \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad
6 0
2 years ago
A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

                            y (x, t) = A*cos(k*x -w*t)  

Where k = stiffness and w = angular frequency

Hence,

                           k = 172 and w = 2730

- Calculate wave speed V:

                            V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

                            T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

                           T = W = Y*V^2 = (w/L*g)*V^2

                            W = (0.0126 / 1.8*9.81)*(13.78)^2

                            W = 0.135 N

4 0
3 years ago
If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?
Lady_Fox [76]
·The acceleration of gravity is proportional to

                       1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is  9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

                                                             =      490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

                 (9.8 m/s²) · (1/5)²  =  0.39 m/s² .

The boy's weight is    (mass) · (gravity)  =  (50kg) · (0.39 m/s²)

                                                                  =     19.6 newtons .

Just as we expected, his weight at that distance is

         (19.6 / 490) = 0.04  =  1/25  =  1/5²  of his weight on the surface.
7 0
2 years ago
Agricultural waste used as a fuel is an example of
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6 0
3 years ago
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What is the relationship between electric field lines and equipotential surfaces?
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An equipotential surface is a three-dimensional version of equipotential lines. * *Equipotential lines are always perpendicular to electric field lines* . The process by which a conductor can be fixed at zero volts by connecting it to the earth with a good conductor is called grounding.
5 0
2 years ago
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