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____ [38]
3 years ago
10

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t

he inner core. The spool is free to roll without slipping on a horizontal surface. If the string unwinds horizontally from the bottom of the core with a constant speed of 25.0 cm/s, what is the speed of the spool
Physics
1 answer:
White raven [17]3 years ago
4 0

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

        v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

         w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

         w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

        \frac{v_o}{r_o} = \frac{v_1}{r_1}

        v1 = \frac{r_1}{r_o} \ \ v_o

let's calculate

       v₁ = \frac{1.50}{1.00} \ \ 25.0

       v₁ = 37.5 cm / s

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kupik [55]
I’m pretty sure it would be 10/20= 0.5m/s2
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A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

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3 years ago
What is Vatiable Velocity?​
Allushta [10]

Answer:

Explanation: When a body covers unequal distances in equal intervals of time in a specified direction, the body is said to be moving with a variable velocity. Example: A rotating fan at a constant speed has variable velocity, because of continuous change in direction.

7 0
2 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
3 years ago
Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it
makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

 Here, we have been given the height through which the ball drops;

 Height of drop = 14.5m - 1.9m  = 12.6m

The right motion equation is;

      V²  = U² + 2gh

V is the final velocity

U is the initial velocity  = 0

g is the acceleration due to gravity  = 9.8m/s²

h is the height

  Now insert the parameters and solve;

       V² = 0² + 2 x 9.8 x 12.6

      V²  = 246.96

       V = √246.96  = 15.7m/s

4 0
3 years ago
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