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mihalych1998 [28]
3 years ago
6

What is the kinetic energy k of the two-car system immediately after the collision? express your answer in terms of ve, vn, and

m?
Physics
1 answer:
AveGali [126]3 years ago
8 0
<span>Consider the following two-caraccident: Two cars of equal mass mcollide at an intersection. Driver E was traveling eastward, anddriver N, northward. After the collision, the two cars remainjoined together and slide, with locked wheels, before coming torest. Police on the scene measure the length d of the skid marks to be 9 meters. The coefficient offriction mubetween the locked wheels and the road is equal to 0.9. Each driver claims that hisspeed was less than 14 meters per second (50 mph). A third driver,who was traveling closely behind driver E prior to the collision,supports driver E's claim by asserting that driver E's speed couldnot have been greater than 12 meters per second. Take the followingsteps to decide whether driver N's statement is consistent with thethird driver's contention. Part A Let the speeds of drivers E and N prior tothe collision be denoted by v_e and v_n, respectively. Find v^2, the square of the speed of the two-car system theinstant after the collision. Express your answer terms ofv_e and v_n. Part B What is the kinetic energy K of the two-car system immediately after thecollision? Express your answer in terms ofv_e, v_n, and m. Part C Write an expression for the work W_fric done on the cars by friction. Express your answer symbolically interms of the mass mof a single car, the magnitude of the acceleration due to gravityg,the coefficient of sliding friction mu, and the distance dthrough which the two-car system slides before coming torest. Part D Using the information given in the problemintroduction and assuming that the third driver is telling thetruth, determine whether driver N has reported his speed correctly.Specifically, if driver E had been traveling with a speed ofexactly 12 meters per second before the collision, what must driverN's speed have been before the collision? Express your answer numerically, inmeters per second, to the nearest integer. Take g, the magnitude of the acceleration due to gravity, to be9.81 meters per second per second.</span>
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3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a
inessss [21]

Answer:

The average angular acceleration is  \alpha =125.487 rad /s^2

Explanation:

From the question we are told that

  From the question we are told that

        The length of the bat is l = 0.85m  \

         The initial linear velocity is  u = 0 m/s

         The time is  t = 0.15s

         The velocity at t is  v = 16 m/s

  Generally average  angular acceleration is mathematically represented as

                \alpha  = \frac{w_f - w_o}{t}

        Where w_f is the finial angular velocity which is mathematically evaluated as  

            w_f = \frac{v}{l}

                  w_f = \frac{16}{0.85}

                        = 18.823 rad/s

 and w_o is the initial angular velocity which is zero since initial linear velocity is zero

               So

                         \alpha  = \frac{18.823 - 0}{0.15}

                               \alpha =125.487 rad /s^2

5 0
3 years ago
The Sun has more gravity than the other planets in our solar system because the Sun is
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B the mass of the solar object determines whether the gravity of the object unless the solar object has Sonic property's such as a neutron star which can be the size of Pluto but have the mass of 900 solar masses
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PLEASEEE HELP, thank you :)
telo118 [61]

Answer:

The answer is B.

Explanation:

Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :

Total resistance = 20Ω + 40Ω + 60Ω

= 120Ω

Next, we have to find out that 1Ω is equal to how many voltage by dividing :

120Ω = 60V

1Ω = 60V ÷ 120

1Ω = 0.5V

Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :

1Ω = 0.5V

20Ω = 0.5V × 20

20Ω = 10V

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