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3 years ago

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A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s. it strikes a second hockey puck with a m

ass 0.11 kg. the first hockey puck comes to rest after the collision. What is the velocity of the second hockey puck after the collision? ( Round your answer to the nearest integer.)

1 answer:

Igoryamba3 years ago

3 0

Answer:

$v_2 = 23.182m/s$

Explanation:

Given

$m_1 = 0.17kg$

$m_2 = 0.11kg$

$u_1 = 15m/s$ -- Initial Velocity of the first

$u_2 = 0m/s$ -- Initial Velocity of the second

$v_1 = 0m/s$ -- Final Velocity of the first

Required

Determine the final velocity of the second hockey puck ( $v_2$ )

This question illustrates law of conservation of momentum and it'll be solved using the following formula:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ ---- law of conservation of momentum

Substitute in the right values

$0.17 * 15 + 0.11*0 = 0.17 * 0 + 0.11*v_2$

$0.17 * 15 + 0 = 0 + 0.11*v_2$

$2.55 = 0.11*v_2$

Solve for v2

$v_2 = 2.55/0.11$

$v_2 = 23.182m/s$

<em>Hence, the final velocity of the second hockey puck is 23.182m/s</em>

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ankoles [38]

Answer:

a. B = 0.20T

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c. E = 0.236J

d. L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

$B=\frac{\mu_o n i}{L}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current = 90.0A

You replace the values of the parameters in the equation (1):

$B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T$

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

$u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}$

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

$E=uV$ (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

$V=AL$

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

$V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3$

Then, the energy contained in the solenoid is:

$E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J$

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

$L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H$

The inductance of the solenoid is 5.84*10^-5 H

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3 years ago

inysia [295]

<u>Hello and Good Morning/Afternoon</u>:

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⇒ must ensure that there is an equal number of elements on both sides of the equation at all times

<u>Let's start balancing:</u>

On the left side of the equation, there are 2 carbon molecule

⇒ but only so far one on the right side

C<em>₂H₅OH + __O₂ → 2CO₂ + __ H₂O</em>

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⇒ but only so far two on the right side

C<em>₂H₅OH + __O₂ → 2CO₂ + 3H₂O</em>

On the right side of the equation, there are 7 oxygen molecules

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C<em>₂H₅OH + 3O₂ → 2CO₂ + 3H₂O</em>

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2 years ago

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Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

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We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

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Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

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3 years ago

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goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

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From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

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