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Andrew [12]
3 years ago
10

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

ultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide
Physics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

The distance covered by puck A before collision is  z = 8.56 \ m

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  v_A =  3.90 \ m/s

        The speed of puck B is  v_B  =  4.30 \ m/s

The distance covered by puck A is mathematically represented as

     z =  v_A * t

  =>  t  =  \frac{z}{v_A}

 The distance covered by puck B  is  mathematically represented as

      18 - z =  v_B  * t

=>   t  = \frac{18 - z}{v_B}

Since the time take before collision is the same

        \frac{18 - z}{V_B}  =  \frac{z}{v_A}

substituting values

          \frac{18 -z }{4.3}  = \frac{z}{3.90}

=>      70.2 - 3.90 z   = 4.3 z

=>       z = 8.56 \ m

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3 years ago
A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a
chubhunter [2.5K]

Answer:

The ballon will brust at

<em>Pmax = 518 Torr ≈ 0.687 Atm </em>

<em />

<em />

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}

Rmax = 10.001 cm

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6 0
3 years ago
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Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

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solving the above equation we get

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the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0
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