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vitfil [10]
3 years ago
12

a student places 8 similar coins in a pile . the height of the pile of coins is 2.4 cm calculate the average thickness of one co

in
Physics
1 answer:
Anon25 [30]3 years ago
7 0

Answer:

0.3 cm

Explanation:

\frac{2.4}{8}  \\  = 0.3

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A snowball accelerates at
ANTONII [103]

Answer:

0. 1226495726kg

Explanation:

Force is the product of mass and acceleration.

Mathematically,

Force(F) = mass (m)×acceleration(a)

Substituting the values into the equation

2. 87=m×23. 4

2. 87=m (23. 4)

2. 87/23. 4=m (23. 4)/23. 4

2. 87/23. 4=m

0. 1226495726=m

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3 years ago
A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.<br><br>What is the friction force?
zmey [24]

Answer:

54.3N

Explanation:

The normal force is perpendicular to the slope, so:

Normal Force = cos(37.2)(9.8*65).......507.39N

F(friction)=mu*F(normal)

F(friction)=(0.107)(507.39)

F(friction)=54.3N

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2 years ago
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Ymorist [56]

Answer: I would say the answer is B.

Explanation: It looks and sounds the most correct.

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3 years ago
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serg [7]
The first diagram is Nuclear Fission which is the splitting of a heavy, unstable nucleus into 2 lighter nuclei.

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3 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

4 0
2 years ago
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