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svet-max [94.6K]
3 years ago
6

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

uniformly at a rate of 5.8 m/s2 for 3.3 seconds. It then continues at a constant speed for 9.8 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 255.79 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.
3)How far does the blue car travel before its brakes are applied to slow down?
Physics
2 answers:
Rudiy273 years ago
7 0
Well, the car gets to 5.8*3.3=19.14 m/s, getting to 5.8*3.3*3.3=63.162 meters from the start. For 9.8 secons it goes 187.572 meters, a total of 250.734. So, in 5.056 meters, it stopped uniformly. The speed dropped from 19.14 to 0 in 5.056 meters. There was a formula which didn't use time, but I've forgotten it. It would have given us the deceleration and then one could find the time through a 2nd degree equation. The timing for the first two parts is already known.
Either way, we've calculate more than we need to know for this question. You can cut the answer from that 250.734 distance.
ioda3 years ago
6 0

Answer:

raheh

Explanation:

zdrhdrh

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If anyone answers this question, i will give you 100 points and give brainliest!
Verizon [17]

Answer:

Q1) C, D

Q2) A

Q3) E

Q4) C

Q5) B

Q6) C

Explanation:

I think these are correct, hopefully it helps!

5 0
3 years ago
A neutron in a nuclear reactor makes an elastic head-on collision with the nucleus of a plutonium atom initially at rest. (a) Wh
olga55 [171]

Answer:

Fraction = 59049/60025

Explanation:

Let m be the mass of the neutron and M be the mass of the plutonium nucleus (at rest)

Now, formula for kinetic energy before collision is;

K_i = ½mu²

Formula for kinetic energy after collision is;

K_f = ½mv²

Where;

u is the velocity of the neutron before collision

v is the velocity of the neutron after collision.

From collision principle where momentum before collision equals momentum after collision, we can say that;

(m - M)u = (m + M)v

Thus,

v = [(m - M)u]/(m + M)

Putting [(m - M)u]/(m + M) for v in the final kinetic energy equation gives;

K_f = ½m([(m - M)u]/(m + M))²

K_f = ½mu²((m - M)²/(m + M)²)

To get the fraction of the neutron's kinetic energy is transferred to the plutonium nucleus, it is simply;

K_f/K_i = [½mu²((m - M)²/(m + M)²)]/½mu²

This gives;

K_f/K_i = ((m - M)²/(m + M)²)

But mass of plutonium = 244m

Thus;

K_f/K_i = ((m - 244m)²/(m + 244m)²)

K_f/K_i = 59049m²/60025m²

K_f/K_i = 59049/60025

4 0
3 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
4 years ago
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I have a bottle of gas, the bottle can expand and contract. Initially the gas is at 1 kpa of pressure and a volume of 1 Liter, a
drek231 [11]

Answer:

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where,

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T₁ = initial temperature = 290 k

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<u>P₂ = 1.22 kPa</u>

7 0
3 years ago
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olasank [31]
<span>Both objects receive the same impulse.</span>
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