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Lapatulllka [165]
3 years ago
12

Which has a larger moment of inertia, a disc of mass 61 kg or a hoop of mass 36 kg with the same radius

Physics
1 answer:
dsp733 years ago
7 0

Answer:

The ring has a larger moment of inertia.

Explanation:

Given that the mass of disc=61 kg

Let the radius of the disc is R

The moment of inertia of the disc having mass m, about the axis passing through the center of the disc,

I_{disc} = \frac {mR^2}{2}=\frac {61R^2}{2}=30.5 R^2

Mass of hoop = 36 kg

As the radius of disc and hoop is the same, so the radius of the hoop is R.

The moment of inertia of the ring (hoop) having mass m, about the axis passing through the center of the disc,

I_{hoop} = mR^2=36 R^2

Hence, the ring has a larger moment of inertia.

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The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0
3 years ago
you are flying on an airplane and you overhear two pasengers comment that flight attendants must be very skiled, one says "when
s344n2d4d5 [400]

How do you get out of a room with only a mirror and a table?

Look in the mirror you see what you saw, use the saw to cut the table, half and half makes a hole, jump in it.

Thought you would like that.

4 0
3 years ago
The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th
MaRussiya [10]

Answer:

a). P=78.4W

b). P=392kPa

c.) It must be at the bottom

Explanation:

Given:

Volume flow V_f=2.0x10^{-4}m^3/s

Well depp h=40.m

a.

The power output of the pum

W=F*d

F=m*g

m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg

W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2

W=78.4J

P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W

b.

The pressure of difference the pum

ΔP=p*g*y'

ΔP=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa

P=392kPa

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way  

4 0
4 years ago
What is solar power good for?
san4es73 [151]

Answer: the earth

Explanation: ask your teacher

3 0
2 years ago
Read 2 more answers
A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
3 years ago
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