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Gemiola [76]
3 years ago
15

Which produces more energy:

Physics
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

a) uranium

b) deuterium

Explanation:

For the first question, you can actually know this by simple logic. The nucleus of Ucranium, has more energy than a pair of deuterium. Ucranium has a mass number and atomic mass higher than deuterium, therefore, when you do a fission with ucranium in it's nucleous, this would be in energy higher than the deuterium.

However, in terms of mass only, the fissioning of 1 gram of Deuterium will produce more energy than 1 gram of uranium, basically because deuterium is lighter and has a lower molecular weight than ucranium. This means that because of this lower molecular weight, in 1 gram of deuterium will contain more number of atoms than in 1 g of uranium. The interaction with these atoms, will produce more energy than the energy produced by uranium.

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azamat
<span>it would bond to the phosphate 

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A person gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. The cart travels for
mash [69]

Answer:

Only a backward force is acting, no forward force.

Explanation:

  • Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
  • As it would move at a constant speed, no net force would be acting on it.
  • So, if it is gradually slowing, there must  be a net force producing an acceleration in a direction opposite to the movement.
  • This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
  • As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
  • This is consistent with the direction of the acceleration of the cart.
8 0
3 years ago
A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter
yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

  • Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

        I_{1} = \frac{V}{R_{int} +r_{L} }

  • where Rint = r and RL = r
  • Replacing these values in I₁, we have:

       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

        \frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}

  • The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.  
7 0
3 years ago
A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

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V=41.96ft/s

b. t=(v-u) /g

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5 0
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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
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Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

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