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Orlov [11]
3 years ago
13

Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m]

, RD= 6 [m], RE=12 [m]. ​

Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

-5 m/s

Explanation:

The linear velocity of B is equal and opposite the linear velocity of E.

vB = -vE

vB = -ωE rE

10 m/s = -ωE (12 m)

ωE = -0.833 rad/s

The angular velocity of E is the same as the angular velocity of D.

ωE = ωD

ωD = -0.833 rad/s

The linear velocity of Q is the same as the linear velocity of D.

vQ = vD

vQ = ωD rD

vQ = (-0.833 rad/s) (6 m)

vQ = -5 m/s

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SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

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<u>D. 475 </u>

Explanation:

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A guitar string is 90 cm long and has a mass of 3.5g . The distance from the bridge to the support post is L=62cm, and the strin
nataly862011 [7]

Answer:

v_1 =  301 Hz

v_2 =  601 \ \ Hz

v_3 =  901 \ Hz

Explanation:

From the question we are told that

     The  length of the string is  l = 90 \ cm  =  0.9 \ m

     The mass of the string is  m_s  =  3.5 \ g =0.0035 \ kg

     The  distance  from the bridge to the support post L =  62 \ c m  =  0.62 \ m

    The tension is T  =  540 \ N

Generally the frequency is mathematically represented as

        v  =  \frac{n}{2 * L }  [\sqrt{ \frac{T}{\mu} } ]

Where n is and integer that defines that overtones

i.e  n =   1 is for fundamental frequency

      n =  2   first overtone

       n =3   second overtone

Also  \mu is the linear density of the string which is mathematically represented as

           \mu  =  \frac{m_s}{l}

=>        \mu  =  \frac{0.0035 }{ 0.9 }

=>       \mu  =  0.003889 \  kg/m

So for   n = 1

     v_1  =  \frac{1}{2 *  0.62 }  [\sqrt{ \frac{ 540}{0.003889} } ]

     v_1  = 301 \ Hz

So for  n =  2

     v_2  =  \frac{2}{2 *  0.62 }  [\sqrt{ \frac{ 540}{0.003889} } ]

     v_2  = 601 \ Hz

So for  n =  3

     v_3  =  \frac{3}{2 *  0.62 }  [\sqrt{ \frac{ 540}{0.003889} } ]

     v  =901  \ Hz

     

       

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