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antoniya [11.8K]
2 years ago
11

(a) Suppose you charge a 2.5 F capacitor with two 1.5 volt batteries. How much charge was on each plate? 7.5 Correct: Your answe

r is correct. C (b) How many excess electrons were on the negative plate? electrons
Physics
1 answer:
natali 33 [55]2 years ago
7 0

Answer:

C = Q / V      

Q = C * V = 2.5 F * 3 V = 7.5 Coulombs (assuming the batteries are in series)

N e = Q       where N is the number of electrons and e the electronic charge

N = 7.5 / (1.60E-19) = 4.7E19 electrons

or 4.7 * 10^19 electrons

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zhuklara [117]

1) 1.08 m/s^2

Explanation:

Acceleration is equal to the change in velocity divided by the time taken:

a=\frac{v-u}{\Delta t}

where

v is the final velocity

u is the initial velocity

\Delta t is the time taken

In this problem, we have:

- initial velocity: u = 0 (you start from rest)

- final velocity: v = 5.4 m/s

- time taken: \Delta t = 5 s

Therefore, the acceleration is

a=\frac{v-u}{\Delta t}=\frac{5.4 m/s-0}{5 s}=1.08 m/s^2


2) -0.54 m/s^2

We can calculate the acceleration to slow down using the same formula as before, but this time the data are as follows:

- initial velocity : u = 5.4 m/s

- final velocity : v = 0 (you come to a stop)

- time taken : \Delta t = 10 s

using the same formula, we find

a=\frac{v-u}{\Delta t}=\frac{0-5.4 m/s}{10 s}=-0.54 m/s^2

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A 0.40-kg mass is attached to a spring with a force constant of k = 277 N/m, and the mass–spring system is set into oscillation
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To solve this problem it is necessary to apply the concepts related to the kinetic energy expressed in terms of simple harmonic movement, as well as the concepts related to angular velocity and acceleration and linear acceleration and velocity.

By definition we know that the angular velocity of a body can be described as a function of mass and spring constant as

\omega = \sqrt{\frac{k}{m}}

Where,

k = Spring constant

m = mass

From the given values the angular velocity would be

\omega = \sqrt{\frac{277}{0.4}}

\omega = 26.31rad/s

The kinetic energy on its part is expressed as

E = \frac{1}{2} m\omega^2A^2

Where,

A = Amplitude

\omega = Angular Velocity

m = Mass

PART A) Replacing previously given values the energy in the system would be

E = \frac{1}{2} m\omega^2A^2

E = \frac{1}{2} (0.4)(26.31)^2(3*10^{-2})^2

E= 0.1245J

PART B) Through the amplitude and angular velocity it is possible to know the linear velocity by means of the relation

v = A\omega

v = (3*10^{-2})(26.31)

v = 0.7893m/s

PART C) Finally, the relationship between linear acceleration and angular velocity is subject to

a = A\omega^2

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When grocery shopping, the mass of the cart changes as you start to fill up your cart. How does the change in mass of your cart
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Answer:

b

Explanation:

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