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nexus9112 [7]
2 years ago
8

Number 18 plz help physics question

Physics
1 answer:
Anvisha [2.4K]2 years ago
8 0

Answer:

Explanation:

a) R = V/I = 0.7/0.022 = 32Ω

b) R = V/I = 0.6/0.005 = 120Ω

c) No...Current change is not linear with voltage change.

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if the velocity of a car is halved the fc required to keeo it in a path of constant radius is multiplied/divided by?
damaskus [11]
I believe this is it
The centripetal force is given by
F = mv^2 / r
When v' = v/2,
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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
2 years ago
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mixas84 [53]

Answer:3.874 m/s^2

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

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34mi/h \approx 15.1994 m/s

we know acceleration is given by =\frac{velocity}{Time}

a=\frac{15.1994-26.8224}{3}

a=-3.874 m/s^2

negative indicates that it is stopping the car

Distance traveled

v^2-u^2=2as

\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s

s=\frac{488.419}{2\times 3.874}

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