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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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A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan
Nesterboy [21]

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

-The quantity of heat is given by multiplying mass by specific heat and by temperature change.

That is; Q = mcΔT

This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.

-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.

7 0
3 years ago
What condition is necessary for an object to make a good reference point?
777dan777 [17]
The object is fixed relative to the motion you are trying to describe.
8 0
3 years ago
Who predict the black hole​
kramer

Answer:

Albert Einstein

Explanation:

He first predicted the existence of black holes in 1916

8 0
2 years ago
Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig
MissTica

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

         Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

         ½ m v₀² + m g y₀ = ½ m v_f²

        v_f² = v₀² + 2 g y₀

         

let's calculate

        v_f = √(10² + 2 9.8 25)

        v_f = 24.3 m / s

6 0
3 years ago
A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibrati
coldgirl [10]

Answer:

1. The period is 1.74 s.

2. The frequency is 0.57 Hz

Explanation:

1. Determination of the the period.

Spring constant (K) = 30 N/m

Mass (m) = 2.3 Kg

Pi (π) = 3.14

Period (T) =?

The period of the vibration can be obtained as follow:

T = 2π√(m/K)

T = 2 × 3.14 × √(2.3 / 30)

T = 6.28 × √(2.3 / 30)

T = 1.74 s

Thus, the period of the vibration is 1.74 s.

2. Determination of the frequency.

Period (T) = 1.74 s

Frequency (f) =?

The frequency of the vibration can be obtained as follow:

f = 1/T

f = 1/1.74

f = 0.57 Hz

Thus, the frequency of the vibration is 0.57 Hz

4 0
3 years ago
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