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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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Explain what voltage,current, resistance are and how they relate to Ohms law. Then give examples
SOVA2 [1]
Voltage is the difference in charge between two points.
Current is the rate the charge flows
Resistance is the tendency a material has to resist the flow of charge (current)
Combining voltage resistance and current Ohm developed the formula
V (Voltage)= I (Current) x R (Resistance)
3 0
3 years ago
Lab number 14: sudden stops hurt newtons first law
kompoz [17]

Answer:

you want to know newtons first law here.

Explanation:

An object that is at rest stays at rest or stays in motion.

7 0
3 years ago
The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a
Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

with m the mass , a the acceleration and \sum\overrightarrow{F} the net force forces that is:

(F-f) (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

(F-f)=ma

solving for a:

a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}

Now let's use the Galileo’s kinematic equation

Vf^{2}=Vo^{2}+2a\varDelta x (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and \varDelta x the time took to achieve that velocity, solving (3) for \varDelta x:

\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}

t=68.94 m

8 0
3 years ago
Why does a black hole have a stronger gravitational pull than the star that collapse to form it?​
Studentka2010 [4]

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

8 0
2 years ago
5. Electromagnetic radiation travels at a speed of 3.0 x 10^8 m/s.
creativ13 [48]

Answer:

i think its b

Explanation:

6 0
2 years ago
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