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3 years ago

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A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s

tationary car that has a perfectly elastic spring bumper. What is the final kinetic energy of the two car system?

1 answer:

Natalka [10]3 years ago

8 0

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

$KE_i = KE_f$

$KE_f = \frac{1}{2} mv^2$

Here,

m = Mass

V = Velocity

Replacing,

$KE_f = \frac{1}{2} (12000)(11)^2$

$KE_f = 72600J$

Therefore the final kinetic energy of the two car system is 72.6kJ

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In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

$^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X$

$^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr$

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

$^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X$

$^{211}_{83}Bi\Rightarrow ^{211}_{84}Po$

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

$^{22}_{11}Na\Rightarrow ^{22}_{11-1}X$

$^{22}_{11}Na\Rightarrow ^{22}_{10}Ne$

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

$^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc$

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

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Mechanical waves propagate or move through a medium because

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The answer to that would be that

they require so its mandatory for mechanical waves to travel through a medium

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The instantaneous velocity of an object is the blank of the object with a blank

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The instantaneous velocity of the object is its speed and direction at that instant.

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Benjamina started her walk from the front door of her ground floor apartment. She walked 6 meters to the corner of the building

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The distance is the total distance she walked which is 16 meters adding the 6 meters to the corner and 10 meters to her friend's apartment. Her displacement is the distance from her original starting point so you set up a triangle with side lengths of 6 and 10 and solve for the hypotenuse which gives you a displacement of 11.66 meters.

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Consider a 40,000 km steel pipe in the shape of a ring that fits snuggly all around the circumference of the Earth. We are heati

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Answer:

The Height is H = 70.02 m

Explanation:

We are given that the

Initial length is = $40000\ Km$ = $40,000 *10^{3} m$

from what we are told in the question the circumference of the circle is = $40,000 Km$

This means that the Radius would be :

Let C denote the circumference

So

$C = 2 \pi r$

=> $r = \frac{C}{2 \pi}$

$r = \frac{40,000}{2 \pi } = \frac{40,000*10^{3}}{2 *3.142} = 6.365*10^6 m$

We are told that 1-meter bar of steel that increases its temperature by 1 degree C will expand $11*10^{-6}$ meters

Hence

The final length would be

$40000*10^3 *(T + \alpha )$

Where T is the change in temperature $\alpha$ is the Coefficient of linear expansion for steel

let $L_{final}$ denote the final length

So

$L_{final} =40000*10^{6} *[1+ 11*10^{-6}]$

$= 40000440 \ m$

Now the Height is mathematically represented as

$Height(H) \ = \frac{change \ in \ radius \ }{2 \pi}$

$= \frac{(40000440-40000*10^3)}{2*3.142}$

$= 70.02m$

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