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astra-53 [7]
3 years ago
15

(b) The figure below shows a model train, travelling at speed , approaching a

Physics
1 answer:
Andrew [12]3 years ago
3 0

Answer:

0.62 m/s

Explanation:

Given that a model train, travelling at speed , approaching a buffer model train buffer spring

The train, of mass 2.5 kg, is stopped by compressing a spring in the buffer.

After the train has stopped, the energy stored in the spring is 0.48 J.

Calculate the initial speed v of the train

Solution

The total energy stored in the spring will be equal to the kinetic energy of the train.

That is,

1/2fe = 1/2mv^2

Substitutes the spring energy, and mass into the formula

0.48 = 1/2 × 2.5 × V^2

2.5V^2 = 0.96

V^2 = 0.96 / 2.5

V^2 = 0.384

V = sqrt ( 0.384 )

V = 0.62 m/s

Therefore, the initial velocity of the train is 0.62 metres per second.

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Two parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively char
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Answer:

K' = 1200 J

Explanation:

To find the kinetic energy you first take into account the formula for the kinetic energy of the charge:

K=\frac{1}{2}mv^2 = 800J   (1)

m: mass of the charge

v: final speed of the charge when it reaches the positively charged plate.

Furthermore, you have that the acceleration of the charge is obtained by using the second Newton law:

F=ma=qE\\\\a=\frac{qE}{m} (2)

a: acceleration

E: electric field

q: charge

The electric field between two parallel plates is V/d, being V the potential difference and d the separation between plates. You replace E in (2) and obtain:

a=\frac{qV}{md}

Next, you take into account the following formula for the calculation of the final speed of the charge:

v^2=v_o^2+2ad\\\\v_o=0m/s\\\\v=\sqrt{\frac{2qVd}{md}}=\sqrt{\frac{2qV}{m}}

Next, you replace this value of v in (1):

K=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{2qV}{m})=qV = 880J   (3)

If the distance between plates is tripled, and the potential difference is halved, you have for the new final speed:

v'^2=v'_o^2+2a(3d)\\\\v_o=0m/s\\\\v'=\sqrt{6ad}=\sqrt{6(\frac{q}{md})\frac{V}{2}d}=\sqrt{\frac{3qV}{m}}

And the kinetic energy becomes:

K'=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{3qV}{m})=\frac{3}{2}qV    (4)

You calculate the ratio between both kinetic energies K and K', that is, you divide equations (3) and (4), in order to find the new kinetic energy:

K=qV=800J\\\\K'=\frac{3}{2}qV\\\\\frac{K}{K'}=\frac{qV}{3/2\ qV}=\frac{2}{3}\\\\K'=\frac{3}{2}K=\frac{3}{2}(800J)=1200J

hence, the kinetic energy of the charge incresases to 1200J

4 0
3 years ago
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
tester [92]

Answer:

7.01yard/sec

Explanation:

Given parameters:

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Unknown:

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Solution:

To solve this problem;

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2) reduces the mass number (protons plus neutrons) by 4. If you start with 239 and subtract two protons and two neutrons, you are left with 235.

Explanation:

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