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3 years ago

Compare and contrast fluorescent and incandescent light bulbs.

1 answer:

5 0

Incandescent light is a glowing white light produced by heat. An incandescent light bulb works by heating a filament in the bulb. Fluorescent light is a bright light produced by electricity flowing through a tube filled with ionized gas. Fluorescent light bulbs are more energy-efficient than incandescent bulbs

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If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T

Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

4 years ago

Define rectilinear propagation of light.??

Tju [1.3M]

Answer:

it relates to the light propensity to travel over one straight line without having any interference in its trajectory

Explanation:

5 0

3 years ago

Read 2 more answers

A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off toge

natka813 [3]

Answer: 909 m/s

Explanation:

Given

Mass of the bullet, m1 = 0.05 kg

Mass of the wooden block, m2 = 5 kg

Final velocities of the block and bullet, v = 9 m/s

Initial velocity of the bullet v1 = ? m/s

From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve

m1v1 = (m1 + m2) v, on substituting, we have

0.05 * v1 = (0.05 + 5) * 9

0.05 * v1 = 5.05 * 9

0.05 * v1 = 45.45

v1 = 45.45 / 0.05

v1 = 909 m/s

Thus, the original velocity of the bullet was 909 m/s

3 0

3 years ago

5. If a spring with a 40 N/m spring constant is used to push a 10 kg ball to a speed of 2 m/s,

kolbaska11 [484]

By compressing the spring a distance x (in m), you are storing 1/2 k x ² (in J) of potential energy, which is converted completely into kinetic energy 1/2 m v ², where

• k = 40 N/m = spring constant

• m = 10 kg = mass of the ball

• v = 2 m/s = ball's speed (at the moment the spring returns to its equilibrium point)

So we have

1/2 k x ² = 1/2 m v ²

x = √(m/k v ²) = √((10 kg) / (40 N/m) (2 m/s)²) = 1 m

3 0

3 years ago