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frez [133]
3 years ago
14

A 27.0 g marble sliding to the right at 56.8 cm/s overtakes and collides elastically with a 13.5 g marble moving in the same dir

ection at 14.2 cm/s. After the collision, the 13.5 g marble moves to the right at 71 cm/s. Find the velocity of the 27.0 g marble after the collision.
Physics
1 answer:
Nataly [62]3 years ago
7 0

Answer:v_1=28.4 cm/s

Explanation:

Given

mass of marble m_1=27 gm

velocity of marble u_1=56.8 cm/s \approx 0.568 m/s

mass of second marble m_2=13.5 gm

Velocity of second marble u_2=14.2 cm/s \approx 0.142 m/s

After collision 13.5 gm marble moves to the right  at i.e. v_2=71 cm/s

Conserving momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

27\times 56.8+13.5\times 14.2=27\times v_1+13.5\times 71

1533.6+191.7=27\cdot v_1+958.5

27\cdot v_1=766.8

v_1=\frac{766.8}{27}=28.4 cm/s

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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)}

t = 83.92 nm

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Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

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