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lbvjy [14]
3 years ago
7

If a substance or object has the ability to do work, then it has

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
7 0
Energy is the ability to do work.
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In thermodynamics, work is typically done by
kondor19780726 [428]
The answer is gases
3 0
3 years ago
Compute the work performed when 32 pounds is lifted 10 feet.
Murljashka [212]
W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
W = 142 N * 3.048 m = 434 J
(I approximated the conversions- I hope it helps)
7 0
3 years ago
I need help with these. Please show workings<br>​
Sauron [17]

Answer:

Imp = 25 [kg*m/s]

v₂= 20 [m/s]

Explanation:

In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.

1)

(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})

where:

m₁ = mass of the object = 5 [kg]

v₁ = initial velocity = 0 (initially at rest)

F = force = 5 [N]

t = time = 5 [s]

v₂ = velocity after the momentum [m/s]

(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]

2)

(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]

8 0
3 years ago
Read 2 more answers
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
3 years ago
A ball hits a wall horizontally at 6m/s and rebounces at 4.4m/s the ball is in contact with wall for 0.04 sec. what is the accel
Svetlanka [38]

Answer:

Acceleration (a) = 40 m/s²

Explanation:

Given:

Initial velocity (u) = 6 m/s

Final velocity (v) = 4.4 m/s

Time taken (t) = 0.04sec

Find:

Acceleration (a) = ?

Computation:

We know that,

⇒ v = u + at

⇒ a = (v - u) / t

⇒ Acceleration (a) = (4.4 - 6) / 0.04

⇒ Acceleration (a) = (-1.6) / 0.04

Acceleration (a) = 40 m/s²

8 0
3 years ago
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