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k0ka [10]
2 years ago
4

The technetium isotope 99Tc is useful in medical imaging, but its short 6.0 h half-life means that shipping it from a source won

't work; it must be created where it will be used. Hospitals extract the 99Tc daughter product from the decay of the molybdenum isotope 99Mo. The 99Mo source has a half-life of 2.7 days, so it must be replaced weekly. A hospital receives a 20 GBq 99Mo source.
What is its activity one week later?

Express your answer using two significant figures.
Physics
1 answer:
Artyom0805 [142]2 years ago
3 0

Answer:

Explanation:

Initial radioactivity =  20 GBq

half life of source Mo = 2.7 days

dN / dt = N₀ λ where dN / dt is initial activity , N₀ is initial number of atoms and λ is disintegration constant .

λ = .693 / half life

= .693 / 2.7

= .257

dN / dt = N₀ λ

20k = N₀ x .257

N₀ = 20 k /.257

N = N_0e^{-\lambda t }

= 20 k /.257  x e^{-.257\times  7 }

= 20k x .64383

dN / dt = N λ

dN / dt = 20k x .64383  x .257

= 3.3 GBq.

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A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude
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The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
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- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2 directed downwards, and with initial velocity v_y=v_= sin 30^{\circ}. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height. 
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2 years ago
A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to
Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

0=y_{0}+\frac{1}{2}at^{2}

and solve for t:

t=\sqrt{\frac{-2y_0}{a}}

now we can substitute the values:

t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the  distance from the ship to point A to be:

x_{A}=V_{x}t

x_{A}=(100m/s)(3.19)

x_{A}=319m

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}

So we can substitute the given values:

x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}

Which yields:

x_{f}=329.18m

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

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2 years ago
A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Delicious77 [7]

Answer:

W = -510.98J

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W = F*d*cosA

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This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating
Olegator [25]

Answer:

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PilotLPTM [1.2K]

As it is given that Bulk modulus  and density related to velocity of sound

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B = \frac{kg}{m* s^2}

SO above is the SI unit of bulk Modulus

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2 years ago
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