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3 years ago

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Light refracts when traveling from air into glass because light Question 1 options: a. travels at the same speed in air and in g

lass b. frequency is greater in air than in glass. c. frequency is greater in glass than in air. d. travels slower in glass than in air.

1 answer:

aalyn [17]3 years ago

3 0

D. Travels slower in glass than in air..bends away from normal

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In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary sec

Pie

Answer:

The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

Explanation:

Given;

mass of the first object, m₁ = 0.3 kg

initial velocity of the first ball, u₁ = 2.8 m/s

mass of the second ball, m₂ = 0.4 kg

initial velocity of the second ball, u₂ = 0

let the final velocity of the first ball, = v₁

let the final velocity of the second ball, = v₂

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂

0.84 = 0.3v₁ + 0.4v₂

2.8 = v₁ + 1.333v₂ -------equation (1)

Apply one-direction velocity;

u₁ + v₁ = u₂ + v₂

2.8 + v₁ = 0 + v₂

v₂ = 2.8 + v₁

substitute the value of v₂ into equation (1)

2.8 = v₁ + 1.333v₂

2.8 = v₁ + 1.333(2.8 + v₁)

2.8 = v₁ + 3.732 + 1.333v₁

2.8 - 3.732 = v₁ + 1.333v₁

-0.932 = 2.333v₁

v₁ = -0.932 / 2.333

v₁ = -0.4 m/s

Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

8 0

3 years ago

An 8000-kg engine pulls a 40,000 kg train along a level track and gives it an acceleration a1 = 1.20 m/s 2 . What acceleration (

mel-nik [20]

a = F/(m1+m2)
1.2 = F/48,000
F = 57,600N

a = F/(m1+m3)
a = 57,600/24,000
a = 2.4 m/s^2

8 0

4 years ago

Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.

romanna [79]

Answer:

(a): $\rm -5.627\times 10^3\ N.$

(b): $\rm 7.626\times 10^2\ N.$

Explanation:

<u>Given:</u>

Charge on one sphere, $\rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.$

Charge on second sphere, $\rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.$
Separation between the spheres, $\rm r=3.59\ cm = 3.59\times 10^{-2}\ m.$

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}$

where,

k is called the Coulomb's constant, whose value is $\rm 9\times 10^9\ Nm^2/C^2.$

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.$

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, $\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.$

The new charges on both the spheres are equal and given by

$\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.$

The magnitude of the force that each sphere now experiences is given by

$\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.$

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When is something weightless

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