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3 years ago

Half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value after

Physics

1 answer:

makkiz [27]3 years ago

3 0

Answer:

44 years

Explanation:

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44

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What can you say about the speed of light in diamond?

Inessa05 [86]

So n=c/v, n= index, c=speed of light and v= speed of light in diamond. 2.42=c/v so v=c/2.42, c≈3x108 m/sec so v=1.24x108 m/sec.
Hope this helps.

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3 years ago

Distinguish between a converging lens and a diverging lens.

andriy [413]

Answer:A converging lens is thickest in the middle and causes parallel light rays to converge through the focal point on the opposite side of the lens. A diverging lens is thinner in the middle and causes parallel light rays to diverge away from the focal point on the same side of the lens.

Explanation:

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4 years ago

Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are

Mars2501 [29]

Answer:

θ = 36°

Explanation:

given,

F₁ = 22.8 N

F₂ = 16.6 N

magnitude of force = ?

direction of force = ?

$F = \sqrt{F_1^2 + F_2^2}$

$F = \sqrt{22.8^2 + 16.6^2}$

$F = \sqrt{795.4}$

F = 28.20 N

direction

$\theta = tan^{-1}(\dfrac{F_2}{F_1})$

$\theta = tan^{-1}(\dfrac{16.6}{22.8})$

$\theta = tan^{-1}(0.728)$

θ = 36°

5 0

3 years ago

What is the flow rate of water in a pipe flowing full with an area of 0.3 m2 and velocity of 2.5 m/s?

sashaice [31]

Answer:

0.75 m³/s

Explanation:

Applying,

Q = vA.................... Equation 1

Where Q = flow rate of the water, v = velocity of the water, A = area of the pipe.

From the question,

Given: v = 2.5 m/s, A = 0.3 m²

Substitute these values into equation 1

Q = 2.5(0.3)

Q = 0.75 m³/s

Hence the flow rate of water in the pipe is 0.75 m³/s

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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