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3 years ago

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Two carts, cart AA (mass 4.00 kgkg) and cart BB (mass 6.50 kgkg) move on a frictionless, horizontal track. Initially, cart BB is

at rest and cart AA is moving toward it at 4.00 m/sm/s. The carts are equipped with ideal spring bumpers, meaning they conserve energy. The collision is head-on, so all motion before and after the collision is along a straight line. Let x x be the direction of the initial motion of cart AA.

1 answer:

tankabanditka [31]3 years ago

3 0

Answer:

$V{_a}'=-0.95m/s$

Explanation:

From the question we are told that:

Mass of cart A $m_a=4kg$

Mass of cart B $m_a=6.50kg$

Speed of cart AA $V-{a}=4.00$

Generally the equation for velocity of A after collision $V{_a}'$ is mathematically given by

$V_{a}'=\frac{M_a-M_b}{M_a+M+b} V_a$

$V{_a}'=\frac{4-6.5}{4+M6.5} *4$

$V{_a}'=\frac{4-6.5}{4+M6.5} *4$

$V{_a}'=-0.95m/s$

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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

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The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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Answer:

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Explanation:

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A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

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Explanation:

Given that,

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Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

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Formula of the acceleration is defined as

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We need to calculate the value of $v\times B$

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topjm [15]

Explanation:

u=166m/s, v=0(at it's highest point final velocity is zero), a=9.8m/s², t=8.6s

by the formula, S=ut+½at².

S=[166×8.6+½.×9.8×(8.6)²]. ...by calculation

S = 1427.6+362.404

S=1790.004m

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