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11Alexandr11 [23.1K]
3 years ago
5

PLZ HELP I DONT GET IT

Physics
1 answer:
GaryK [48]3 years ago
6 0

Answer:

0.1 L

Explanation:

From the question given above, we obtained the following data:

Initial volume (V₁) = 0.05 L

Initial Pressure (P₁) = 207 KPa

Final pressure (P₂) = 101 KPa

Final volume (V₂) =?

We can obtain the new volume (i.e the final volume) of the gas by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

207 × 0.05 = 101 × V₂

10.35 = 101 × V₂

Divide both side by 101

V₂ = 10.35 / 101

V₂ = 0.1 L

Thus, the new volume of the gas is 0.1 L

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Let's say the pitcher is angry or just has a really heavy hand while throwing this ball, and now you have to catch it, otherwise it's going to ram into your face. When you put your hands up just in time to catch this ball, this is called impulse, or commonly expressed as a reflex. Depending on what kind of ball is being thrown, such as a golf ball, baseball, basketball, beach-ball, rubber-ball, baseball, etc. ... the weight of the ball itself is going to impact how much it i going to hurt when you catch it without any hand protection. However, if you're catching, let's say a baseball, with a padded glove, it is not going to hurt as bad as catching the baseball bare handed, because the padded glove has enough padding in it to create a barrier between the skin of your hand and the palm of the glove.

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A worker uses a cart to move a load of bricks weighing 680 N a distance of 10 m across a parking lot.
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What is the Binary compound for KP?
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An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a poi
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Answer:

B= 55.6×10^(-7) Tesla

Explanation:

B= μoI/(2πr)

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A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

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At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

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