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Nana76 [90]
3 years ago
7

What is the net charge on a sphere that has the following? (a) 5.87 106 electrons and 8.11 106 protons C

Physics
2 answers:
hammer [34]3 years ago
6 0

Answer:

3.58*10⁻¹³ C

Explanation:

When we have a net charge on a sphere, this means that there must be a difference in the number of electrons and protons on the sphere, otherwise, the sphere would be electrically neutral.

In this case, we can find this difference as follows:

Np -Ne = 8.11*10⁶ protons - 5.87*10⁶ electrons = 2.24*10⁶ protons.

The total charge carried by all these protons is just the product of the charge of only one proton (which is equal to the elementary charge e) times the number of excess protons, as follows:

Qnet = 2.24*10⁶ protons * 1.6*10⁻19 C/proton = 3.58*10⁻¹³ C

AysviL [449]3 years ago
3 0

Answer:

3.6 * 10^-13 C

Explanation:

The net charge of the sphere will be the sum of the total electron charge and total proton charge. Mathematically,

Q = Qe + Qp

TOTAL ELECTRON CHARGE:

An electron has an electronic charge of -1.6022 * 10^-19C.

Hence, the charge of 5.87 * 10^6 electrons will be:

Qe = - 1.6022 * 10^-19 * 5.87 * 10^6

Qe = - 9.405 * 10^-13 C

TOTAL PROTON CHARGE:

A proton has an electronic charge of 1.6022 * 10^-19. Hence, 8.11 * 10^6 protons will have:

Qp = 1.6022 * 10^-19 * 8.11 * 10^6

Qp = 1.3 * 10^-12 C

=> Q = (-9.405 * 10^-13) + (1.3 * 10^-12)

Q = 3.6 * 10^-13 C

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Answer:

S = V t     where S is the horizontal distance traveled

1/2 g t^2 = H       where H is the vertical distance traveled

t^2 = 2 H / g

V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations

tan theta = H / S

V^2 = S g / (2 tan theta)

Using S = L cos theta

V^2 = L g cos theta / (2 tan theta)

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2 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
A piece of gym equipment states that the maximum load it can hold is 300 kg. Why do you think it is important not to go over thi
Crazy boy [7]

Answer:

The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.

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2 years ago
Select all the correct locations on the image. Select the volume units that are greater than one liter.
dalvyx [7]

Answer:

N/A

Explanation:

No Image provided

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A car starts from rest and reaches a velocity of 50.0 m/s in 41.0 seconds. What is the
Nataly_w [17]

Answer:

1.21 or 1.2 acceleration

Explanation:

i think it is 50.0/41.0

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