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Iteru [2.4K]
2 years ago
10

A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s

ee below). One end of the string is held fixed in space. If the cylinder falls as the string unwinds without slipping, what is the acceleration of the cylinder? (Enter the magnitude. Use the following as necessary: g, m, and r.)
Physics
1 answer:
lbvjy [14]2 years ago
4 0

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2

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A sound wave propagates through a region filled with an ideal gas at constant temperature T. It approaches an acoustically perme
nataly862011 [7]

Answer:

The answer is <em>e.2</em>

Explanation:

We should make use of Snell's refractive law. The arriving wave has a certain velocity at T in a medium, then instantly it reaches a medium (same composition) at T' where velocity would either decrease or increase.

When the incidence angle is 30 °, and we want to make the refraction angle 90 ° such that no sound passes through the barrier (this would be named total internal refraction), so we want the second medium to be "faster" than in the first.

<em>The steps are in the image attached:</em>

5 0
2 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
3 years ago
A rod of very small diameter with a mass 2m and length 3L is placed along the xaxis with one end at the origin. An identical rod
rewona [7]

Answer:

coordinates of the center of mass for these two rods

(x_{cm}, y_{cm})= (\frac{3L}{4},  \frac{3L}{4})cm

Explanation:

given

mass of a rod = 2m

length of the rod = 3L

mass of two rods = 2(2m) = 4m

radius = diameter/2 = \frac{3L}{2}

attached is the diagram and solution to the question

5 0
3 years ago
The length of vector A is 93.8 meters and the length of Ay is 38.4 meters, then the length of A, must be
vaieri [72.5K]

Answer:

85.556metres

Explanation:

Using pythagorean theorem

C²=A²+B²

we have c as the hypotenuse vector A thus:

93.8²=A²+38.4²

93.8²-38.4²=A²

8794.44-1474.56=A²

7319.88=A²

A=85.556

3 0
2 years ago
I NEED THIS ASAP!!! A ball is thrown straight up with an initial velocity of 4.40 m/s. Assuming there is no air friction, what i
Lilit [14]

Answer:

I think it is 80m/s

Explanation:

d = ½ g t2

  = ½ (10 m/s2) (4 s)2

  = (5 m/s2) (16 s2)

  = 80 ms

75% sure

Hope this helps!!!

5 0
2 years ago
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