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4 years ago

9

Based on the information in the graph which isotope is the most stable and would therefore not release energy by either fusion o

r fission?
A. Iron-56(Fe-56)
B. Uranium-235(U-235)
C. Hydrogen-2(H-2)

1 answer:

Crank4 years ago

3 0

Answer: The answer is A just answered it

Explanation:

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PLEASE HELP !! ILL GIVE BRAINLIEST A projectile is launched from the top of a 15 m tall building at a speed of 25ms at an angle

mihalych1998 [28]

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3 years ago

Read 2 more answers

Two Objects of the same size will always have the same mass” Is this statement correct?

attashe74 [19]

No, they won't, mass coincides with density and objects have different densities a one pound lead ball would be smaller than a one pound copper one.

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3 years ago

What happens when a wave encounters a boundary and then goes back to the original medium?

lawyer [7]

It causes a movement of convection in the water resulting in a pulling current.

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3 years ago

A 1-m3 tank containing air @ 25 oC & 500 kPa is connected to another tank containing 5 kg of air at 35 oC & 200 kPa thro

VladimirAG [237]

Answer:

Volume of Tank 2, V' = $2.17 m^{3}$

Equilibrium Pressure, $P_{eq} = 278.82 kPa$

Given:

Volume of Tank 1, V = 1 $m^{3}$

Temperature of Tank 1, T = $25^{\circ}C$ = 298 K

Pressure of Tank 1, P = 500 kPa

Mass of air in Tank 2, m = 5 kg

Temperature of tank 2, T' = $35^{\circ}C$ = 303 K

Pressure of Tank 2, P' = 200 kPa

Equilibrium temperature, $20^{\circ}C$ = 293 K

Solution:

For Tank 1, mass of air in tank can be calculated by:

PV = m'RT

$m' = \frac{PV}{RT}$

$m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg$

Also, from the eqn:

PV' = mRT

V' = volume of Tank 2

Thus

V' = $\frac{mRT}{P}$

V' = $\frac{5\times 0.287\times 303}{200} = 2.17 m^{3}$

Now,

Total Volume, V'' = V + V' = 1 + 2.17 = 3.17 $m^{3}$

Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg

Final equilibrium pressure, P'' is given by:

$P_{eq}V'' = m''RT_{eq}$

$P_{eq} = \frac{m''RT_{eq}}{V''}$

$P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa$

$P_{eq} = 287.82 kPa$

3 0

3 years ago

A shark swims around his circular pool. What is the centripetal force of the 681kg shark if the pool has a diameter of 14m and i

blsea [12.9K]

<span>29 Newtons
Centripetal force is modeled by the equation:
F = MV^2/R where
F = force
M = Mass
V = Velocity
R = Radius

I will assume the shark is swimming in a circle with a radius of 14 m / 2 = 7 m. The actual radius will be smaller, but I'll assume that the question is simply poorly worded. So we need to calculate the velocity of the shark. The circumference of the circle is pi*14m = 43.98229715 m. And since velocity is defined as distance over time, we have a velocity of 43.98229715 m / 80 s = 0.549778714 m/s.

Now substitute the known values into the equation for centripetal force.
F = MV^2/R
F = 681 kg * ( 0.549778714 m/s)^2 / 7m
F = 681 kg * 0.302256635 m^2/s^2 / 7m
F = 205.8367683 kg*m^2/s^2 / 7m
F = 29.40525261 kg*m/s^2
F = 29.40525261 N
Rounding to 2 significant figures gives a force of 29 Newtons.</span>

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4 years ago