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padilas [110]
3 years ago
6

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?
Physics
1 answer:
dsp733 years ago
4 0

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

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Crude oil is a mixture of many different components. The extraction of crude oil from the Earth is important, but its refinement
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3 years ago
If 270 watts of power is used in 42 seconds, how much work was done<br>​
navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

W = 270(42)

<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0
3 years ago
Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

       L =   μ₀ n r / 2I

4 0
3 years ago
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