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3 years ago

What term defines the specific amount of time required for half of a radioactive substance to become stable?

Physics

1 answer:

Tanya [424]3 years ago

5 0

A would be the answer because thats how long it takes

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Two students walk in the same direction along a straight path at a constant speed—one at 0.90 m/s and the other at 1.90 m/s.

lakkis [162]

I think the answer to your question is a

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4 years ago

the public is not yet able to purchase hydrogen fuel cell powered cars because engineers have to determine

grigory [225]

Answer: This would more a practical feedback than scientific.

Explanation: Despite of the protocols and processes scientists need to develop in order to stablish a new source of energy or daily life option, logistic issues are one of the most important things on the "to do list".

Would it be very expensive to acquire this kind of fuel? Is it efficient? Is it dangerous to human life or the environment? What about the negative reactions this brand new fuel could have in society?.

3 0

4 years ago

Help cant figure out which ones right

Ahat [919]

D.) It is an "Element".

[ Element cannot be separated by any means ]

Hope this helps!

4 0

4 years ago

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo

Sonbull [250]

Answer:

$2.267$

Explanation:

Drag force is given by

$F=\dfrac{1}{2}\rho Av^2C$

C = Drag coefficient is constant

A = Area is constant

$v_1$ = Velocity of the passenger jet = 1200 km/h = $\dfrac{1200}{3.6}\ \text{m/s}$

$v_2$ = Velocity of the prop plane = $\dfrac{1}{4}v_1$

$\rho_1$ = Density of the air where the jet was flying = $0.38\ \text{kg/m}^3$

$\rho_2$ = Density of the air where the prop plane was flying = $0.67\ \text{kg/m}^3$

$F\propto \rho v^2$

$\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267$

The ratio of the drag forces is $2.267$ .

5 0

3 years ago

Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2, đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g,

dusya [7]

Answer:

The angle is 18.3 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?

surface charge density, σ = 4 x 10^-5 C/m^2

Charge, q = 10^-8 C

mass, m = 0.008 kg

Let the angle is A and the tension in the string is T.

The electric field due to a plane is

$E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\$

Now equate the forces,

$T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree$

5 0

3 years ago