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3 years ago

HELLLPPP why didnt people in marie curies time realize radioactivity could be harmfull?

Physics

2 answers:

RUDIKE [14]3 years ago

7 0

Answer:

cause they were dumb

Explanation:

Anika [276]3 years ago

5 0

Answer:Health

Throughout her work with radium, Marie was unaware of the effects of radioactivity exposure on the body. In her lab, she would keep tubes of radium in her pocket. [3] She began to suspect that radium negatively impacted health when one of her fellow researchers died of a blood disease, and then a few years later her personal assistant died of a blood disease. Even though she suspected that radium exposure was bad for her health, she did very little to monitor her own blood. In 1932, she broke her wrist and the break took much longer to heal than it should have. She then began to notice that her vision was deteriorating and radiation burns on her fingers were becoming more and more painful. Some days she felt too ill to even go to the lab, and finally on July 4, 1934, Marie died from aplastic anemia. [1]

Radium Exposure Treatments

Marie suspected that her health was being negatively impacted by radium exposure, but did nothing about it, most likely because there weren't any effective treatments for radium poisoning yet. At the time, scientists knew that radium was metabolized like calcium. In an attempt to remove it from the system, they manipulated calcium intake. [4] This caused little to no improvements, so parathyroid hormone was added to the treatment. Again, there was some reduction of radium, but not a significant amount. It wasn't until after Marie's death that they realized once radium is in the bones, it is extremely difficult to extract. The lack of therapies for radium exposure may explain why Marie just ignored her symptoms, because she was fully aware of her fate. [4]

© Jenna Gray. The author warrants that the work is the author's own and that Stanford University provided no input other than typesetting and referencing guidelines. The author grants permission to copy, distribute and display this work in unaltered form, with attribution to the author, for noncommercial purposes only. All other rights, including commercial rights, are reserved to the author.

Explanation:

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Ivenika [448]

In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

W = P(V2 - V1)
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3 years ago

A manufacturer selected a metal to use in producing a lightweight button for clothing. A metal that has a density of 2.71 g/cm3

Natali5045456 [20]

Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:

Metal 1

ρ = m/V

ρ = 22.1 g / 3 cm³

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Metal 2

ρ = m/V

ρ = 42 g / 4 cm³

ρ = 10.5 g / cm³

Metal 3

ρ = m/V

ρ = 9.32 g / 5 cm³

ρ = 1.864 g / cm³

Metal 4

ρ = m/V

ρ = 8.13 g / 3 cm³

ρ = 2.71 g / cm³

<h2>R / Metal 4 was selected.</h2>

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2 years ago

Do magnets have to touch each other in order to experience a magnetic force

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3 years ago

Read 2 more answers

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the

Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

$W_{r} = |F_{r}|*|d|*cos(\theta_{1})$

Where:

$F_{r}$ : is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

$W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J$

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

$W_{g} = |F_{g}|*|d|*cos(\theta_{2})$

The force of gravity is given by the weight of the bike.

$F_{g} = -mgsin(24)$

And the angle between the force of gravity and the direction of motion is 180°.

$W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})$

$W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J$

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.

I hope it helps you!

3 0

2 years ago

HELLLPPP why didnt people in marie curies time realize radioactivity could be harmfull?​

HELLLPPP why didnt people in marie curies time realize radioactivity could be harmfull?