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astra-53 [7]
3 years ago
9

PLEASE HELPPPPP!!!! :)

Physics
1 answer:
GuDViN [60]3 years ago
4 0

Answer:

F=X.F=mxq. 3. 1 N/kg=0.5kg g=9.80

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D. The noise of a car horn. Hope this helps! Please comment if it does!
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Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod
Dima020 [189]

Answer:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>

Explanation:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.

<u>We can prove this by the equation of heat for the two bodies:</u>

<em>According to given condition,</em>

\Delta T_1=\Delta T_2

\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}

<em>when there is no heat loss from the system of two bodies then </em>Q_1=Q_2

\frac{1}{m.c_1} =\frac{1}{m.c_2}

\Rightarrow c_1=c_2

  • Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

  • when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

OR

  • the mass of colder object is half the mass of the hotter object while their specific heat is same.
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Which of the following equations illustrates the law of conservation of
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I think the answer is letter B
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okay

Explanation:

6 0
3 years ago
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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
3 years ago
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