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lys-0071 [83]
3 years ago
13

If the half-life of a 2.0 gram sample of a radionuclide is 15 hours, then the half-life of a 1.0 gram sample of the same radionu

clide would be
Physics
1 answer:
Lerok [7]3 years ago
5 0

7.5 hours or 450 minutes. 15/2=7.5

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A scientist is trying to determine the identity of an element. It is highly reactive to water and forms an ionic bond with chlor
wel

The answer is noble gas. Since noble gas are constant and unreactive. They can still shape compounds with other elements. 
Group 15 is also group 5A and Group 17 is also group 7A. Elements in these sets do not typically form ionic bonds; they are more on creating covalent bonds since they're non-metals. 
Therefore, that leaves us with B. from Group 1. They are metals (but Hydrogen) which respond violently with water, and they form ionic bonds, for they drop outer electrons easily.

8 0
4 years ago
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

where k = 9*10^9Nm/C is the coulomb's constant.

Now, in our case

R = 0.0075m

Q = 0.0035C;

therefore,

E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
3 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
3 years ago
While looking at bacteria under a microscope you decide to take a break when u return the amount of bacteria has doubled in size
VikaD [51]
Your break probably only took 20-30 minutes. Bacteria can double every 20-30 minutes. Hope this helped!
~ Sincerely,
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XOXO
4 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
4 years ago
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