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aleksklad [387]

3 years ago

8

Erica (38 kgkg ) and Danny (46 kgkg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is

moving upward past her at 4.9 m/sm/s . At that instant he grabs hold of her.

1 answer:

salantis [7]3 years ago

6 0

Answer:

The speed with which just after he grabs her is 2.68 m/s.

Explanation:

Given that,

Mass of Erica, m = 38 m/s

Mass of Danny, m' = 46 kg

Erica reaches the high point of her bounce, Danny is moving upward past her at 4.9 m/s. At this moment, the initial speed of Erica will be 0. The momentum will remain conserved. Using the conservation of linear momentum as :

$mu+m'u'=(m+m')V\\\\0+46\times 4.9=(38+46)V\\\\V=\dfrac{225.4}{84}\\\\V=2.68\ m/s$

So, the speed with which just after he grabs her is 2.68 m/s. Hence, this is the required solution.

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2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

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Answer:

a

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Stranded Wire $I_2 = 0.00978 \ A$

b

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Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

2 years ago

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s

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Answer:

a

$\theta = 0.0022 rad$

b

$I = 0.000304 I_o$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

The distance of the slit separation is $d = 0.500 \ mm = 5.0 *10^{-4} \ m$

Generally the condition for two slit interference is

$dsin \theta = m \lambda$

Where m is the order which is given from the question as m = 2

=> $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

substituting values

$\theta = 0.0022 rad$

Now on the second question

The distance of separation of the slit is

$d = 0.300 \ mm = 3.0 *10^{-4} \ m$

The intensity at the the angular position in part "a" is mathematically evaluated as

$I = I_o [\frac{sin \beta}{\beta} ]^2$

Where $\beta$ is mathematically evaluated as

$\beta = \frac{\pi * d * sin(\theta )}{\lambda }$

substituting values

$\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }$

$\beta = 0.06581$

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$I = I_o [\frac{sin (0.06581)}{0.06581} ]^2$

$I = 0.000304 I_o$

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2 years ago