Physics, 27.05.2020 06:57 harleyloveless2853

Help!!
A 0.368 kg bead can slide on a curved wire as seen in the figure.
Assume h1 = 4.20 m and h2 = 2.14 m. If the wire is frictionless and the bead is released with an initial speed of 1.96 m/s in the forward direction then find the speed of the bead at B.
If the wire is frictionless, find the speed of the bead at C.

Help!! A 0.368 kg bead can slide on a curved wire as seen in the figure. Assume h1 = 4.20 m and h2

Answers: 3

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Answer from: irickkadenl

The velocity at B is $v = 12.6 \ m/s$

The velocity at C is $v_c =10.80 \ m/s$

Explanation:

From the question we are told that

The mass of the bead is $m_b = 0.368 \ kg$

The first height is $h_1 = 4.20 \ m$

The second height is $h_2 = 2.14 \ m$

The initial speed of the bead is $u = 1.96 \ m/s$

According to the law of energy conservation

KE _ A + PE_A = KE_B + PE_B

Now KE _ i is the kinetic energy at A and it is mathematically represented as

$KE_A = \frac{1}{2} mu^2$

PE_i is the potential energy at A which is mathematically represented as

PE_A = mg h_1

KE_f is the kinetic energy at B which is mathematically represented as

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

Where v is the speed of the bead at B

PE_B is the potential energy at B which equal to 0 because height is 0 at B

$\frac{1}{2} mu^2 + mg h_1 = \frac{1}{2} mv^2$

Making v the subject

$v = \sqrt{u ^2 + 2gh_1}$

substituting values

$v = \sqrt{1.96 ^2 + 2* 9.8 * 4.20}$

$v = 12.6 \ m/s$

According to the law of energy conservation

KE _ B + PE_B = KE_C + PE_C

$KE_B = \frac{1}{2} m v^2$

PE_B = 0

KE_C is the kinetic energy at c which is mathematically represented as

$KE_C = \frac{1}{2} * m v_c^2$

PE_C is the potential energy at C which is mathematically represented as

PE_C = mg h_2

$\frac{1}{2} m v^2 = \frac{1}{2} * m v_c^2 + mg h_2$

v^2 = v_c^2 + 2g h_2

making v_c the subject

$v_c = \sqrt{12.6^2 - 2* 9.8 * 2.14}$

$v_c =10.80 \ m/s$

Answer from: Quest

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

$v_j = v_1 + v_2 = 3.100 + 1.350 = 4.45 m/s$

now the speed of paul will be same as it is given as he is running on the ground

$v_p = 3.100 m/s$

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

$\delta t = \frac{d}{v_j} - \frac{d}{v_p}$

$\delta t = \frac{2080}{4.450} - \frac{2080}{3.100}$

$\delta t = -203.55 s$

so jerome will reach first and the time difference will be 203.55 s or 3.4 min

Answer from: Quest

the journeys of marquette, joliet and la salle a) new france expand.

all three of these explorers explored the midwest and the area around the great lakes, which they subsequently claimed for france.

this was a big portion of territory and expanded new france which before then only consisted of quebec province in northeastern canada

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Help!!
A 0.368 kg bead can slide on a curved wire as seen in the figure.
Assume h1 = 4.2...

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