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3 years ago

If two objects collid and one is initially at rest, is it possible for both to be at rest after the collision?

Physics

1 answer:

Ludmilka [50]3 years ago

5 0

Answer:

Is it NOT possible for both particles to be at rest after the collision

Explanation:

Law Of Conservation Of Linear Momentum

The total linear momentum of a system of particles is conserved regardless of their internal interactions while no external force is applied to the system. If $m_1$ and $m_2$ are the masses of two particles, and only one of them is at rest, there is a non-zero total linear momentum, i.e.

$P_1=m_1v_1+m_2v_2$

if at least one of the speeds is not zero, the total initial linear momentum is not zero.

If after the collision, both particles remain at rest, both speeds will be zero and the total linear momentum won't be conserved.

So, is it NOT possible for both particles to be at rest after the collision

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Calculate the minimum thickness (in nm) of an oil slick on water that appears blue when illuminated by white light perpendicular

mr_godi [17]

Answer:

The minimum thickness = 83.92 nm

Explanation:

The relation between the wavelength in a particular medium and refractive index $\lambda_n = \frac{ \lambda }{n}$

where ;

$\lambda$ = wavelength of the light in vacuum

n = refractive index of medium with respect to vacuum

For one phase change :

$2t = \frac{\lambda_n}{2}\\\\where \ \lambda_n = \frac{\lambda}{n}\\\\Then \ \\\\2t = \frac{\lambda}{2n}\\\\t = \frac{\lambda_n}{4n}$

Replacing 1.43 for n and 480 nm for λ; we have:

$t = \frac{480}{4(1.43)}$

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3 years ago

What is the weight in newtons of a 10 kg mass on the earth's surface?

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Answer:

98 kg

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Explanation:

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3 years ago

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A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec$

Maximum acceleration is given by $a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2$

So maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

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3 years ago

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2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

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From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

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Romashka-Z-Leto [24]

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