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3 years ago

Based on the information in the graph, why is energy released during the fusion of hydrogen (H) into helium (He)?

Physics

1 answer:

MrMuchimi3 years ago

7 0

Answer:

Explanation:

Just had the same question and couldnt find and answer so i guessed. heres the answer no lol!

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When connected to a battery, a lightbulb glows brightly. If the battery is reversed and reconnected to the bulb, the bulb will g

BartSMP [9]

If the battery is reversed and reconnected to the bulb, the bulb will glow <span>with the same brightness. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that you were looking for and it has actually come to your help.</span>

5 0

4 years ago

Charge A is sitting in an electric field you know the following information:________

soldi70 [24.7K]

Answer:

The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:

∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)

Explanation:

The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.

6 0

3 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

3 years ago

A solid sphere with a radius of 50.0 cm has a total positive charge of 40.0 μC uniformly distributed in its volume. Calculate th

il63 [147K]

Answer:

$E=2.88\times 10^5\ N/C$

Explanation:

It is given that,

The radius of the solid sphere, R = 50 cm = 0.5 m

Charge on the sphere, $q=40\ \mu C=40\times 10^{-6}\ C$

We need to find the magnitude of the electric field r = 10.0 cm away from the center of the sphere. The electric field at point r away form the center of the sphere is given by :

$E=\dfrac{kqr}{R^3}$