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2 years ago

What do nuclear scientists call a stick of uranium?

2 answers:

6 0

They call it "rod" or "fission rod". This is because the uranium is the element used for the nuclear fission, where the uranium nuclei disintegrate producing lighter nuclei and energy.

mihalych1998 [28]2 years ago

5 0

Uranium is material that has 92 protons and 92 electrons with a valence of 6. The chemical is highly used in the nuclear fission for creating nuclear weapons and ends up as radioactive waste which got its own issues of disposal. As it is used in nuclear reactor, nuclear scientist calls the stick of uranium as a fission rod.

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An artificial satellite in a low orbit circles the earth every 98 minutes. what is its angular speed in rad/s?

postnew [5]

To finish one orbit it will take 98 x 60 seconds. So; (2 x pi)/(98 x 60) = 1.07 x 10^-3 rad/sec.

8 0

2 years ago

Which factors could be potential sources of error in the experiment? check all that apply.

Vadim26 [7]

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

6 0

2 years ago

Read 2 more answers

A particle moving in the x direction is being acted upon by a net force F(x)=Cx2, for some constant C. The particle moves from x

elixir [45]

Answer:

Change in kinetic energy is ( 26CL³)/3

Explanation:

Given :

Net force applied, F(x) = Cx² ....(1)

Displacement of the particle from xi = L to xf = 3L.

The work-energy theorem states that change in kinetic energy of the particle is equal to the net amount of work is done to displace the particle.

That is,

ΔK = W = ∫F·dx

Substitute equation (1) in the above equation.

ΔK = ∫Cx²dx

The limit of integration from xi = L to xf = 3L, so

$\Delta K=\frac{C}{3}(x_{f} ^{3} - x_{i} ^{3})$

Substitute the values of xi and xf in the above equation.

$\Delta K=\frac{C}{3}((3L) ^{3} - L ^{3})$

$\Delta K=\frac{C}{3}\times26L^{3}$

5 0

2 years ago

A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Le

Katyanochek1 [597]

Answer:

$\Delta KE = 4.20\times 10^{13}\ J$

Explanation:

given,

mass of spaceship(m) = 8600 Kg

Mass of earth = 5.972 x 10²⁴ Kg

position of movement of space ship

R₁ = 7300 Km

R₂ = 6700 Km

the kinetic energy of the spaceship increases by = ?

Increase in Kinetic energy = decrease in potential energy

$\Delta KE = GMm (\dfrac{1}{R_2}-\dfrac{1}{R_1})$

$\Delta KE = GMm (\dfrac{R_1-R_2}{R_2R_1})$

$\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{7300 - 6700}{7300 \times 6700})$

$\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{600}{48910000})$

$\Delta KE = 4.20\times 10^{13}\ J$

5 0

2 years ago

Assuming the pick-up trucks, trailers and road conditions are exactly the same, which vehicle will take a longer distance to sto

mars1129 [50]

I’ve answered this before so I know the question is missing an important given and that given is: 1 has an empty trailer and the other has a fully loaded one.

So, it would be the fully loaded trailer that would take a longer distance to stop because a lot of weight is being pulled, and when the brakes are started, the fully loaded trailer is more like pushing against the truck.

6 0

1 year ago