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Grace [21]
3 years ago
11

A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare

s the objects' changes in momentum?
A. The bug's change in momentum is larger than the car's change in momentum.
B. The bug's change in momentum is smaller than the car's change in momentum.
C. The bug's change in momentum is equal to the car's change in momentum.
D. The question does not provide enough information. (The question should include whether the collision is elastic or inelastic.)
Physics
1 answer:
zvonat [6]3 years ago
6 0

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

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2 years ago
Nuclear chain reactions within power plants do NOT produce bomb-like explosions primarily because the ________.
Gelneren [198K]

Answer: D. Density of uranium within nuclear fuel rods is insufficient to become explosive

Explanation: Nuclear power plants use the same fuel as nuclear bombs, i.e. radioactive Uranium-235 isotope. However, in a nuclear power plant, the energy is released more slowly unlike in a nuclear bomb. <em>The energy released is through nuclear fission, and radioactive decay occurs at the same rate as in nuclear bombs. therefore, option A, B</em><em> </em><em>and C are incorrect.</em>

The primary reason why nuclear chain reactions within power plants do NOT produce bomb-like explosions is because the uranium fuel rods used in electricity generation is not sufficiently enriched in Uranium-235 to produce a nuclear detonation. This is the same idea in option D which is the correct option.

6 0
3 years ago
if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given
yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given,  y=80t-16t^2 gives the height in 't' seconds.

Average velocity = Rate of change of distance

                             = Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1)  t = 0.01 s

    y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

    Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

    y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

    Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by  y=80t-16t^2 .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

Learn more about average velocity at brainly.com/question/6504879

#SPJ4

7 0
2 years ago
A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane
kondor19780726 [428]

The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

<h3>How to find the final temperature of the gas?</h3>
  • Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.
  • In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.
  • The membrane is raptured without applying any external force, thus, dW=0.
  • We have the first law of thermodynamic expression as,

                                dU=dQ-dW

  • Here it is zero.

                                  dU=0,

  • As we know that,

                             dU=C_pdT=0\\\\thus,  dT=0\\\\or , T=constant\\\\i.e, T_1=T_2

  • Thus, the final temperature of the system will be equal to the initial temperature,

                          T_1=T_2=100^0C=373K

<h3>How much work is done?</h3>
  • We found that the process is isothermal,
  • Thus, the work done will be,

                               W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J

Where, R is the universal gas constant.

<h3>What is a reversible process?</h3>
  • Any process which can be made to proceed in the reverse direction is called reversible process.
  • During which, the system passes through exactly the same states as in the direct process.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

Learn more about thermodynamic processes here:

brainly.com/question/28067625

#SPJ1

7 0
1 year ago
Read 2 more answers
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