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In-s [12.5K]
2 years ago
13

Two long straight wires enter a room through a window. One carries a current of 2.9A into the room, while the other carries a cu

rrent of 4.4 out of the room. Calculate the magnitude (in T.m) of the path integral of B.ds around the window frame.
Physics
1 answer:
Degger [83]2 years ago
4 0

Answer and Explanation:

curents i = 2.9 A

           i ' = 4.4 A

the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed

          = μo* ( i '- i )

Since from Ampere's law

where μ o = permeability of free space = 4π * 10 ^-7 H / m

plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)

                                 = 1.884 * 10^-6 Tm

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A microwave oven operates at 2.60 ghz . what is the wavelength of the radiation produced by this appliance?
cestrela7 [59]
<span>10.3 cm The wavelength will be the distance that light travels in 1 second divided by the frequency of the radiation. Since the over operates at 2.60 ghz, the frequence is 2.6 billion times per second, or 2.60 x 10^9. The speed of light is defined as 299792458 m/s exactly. So 299792458 m/s / 2.60 x 10^9 1/s = 0.10337671 m = 10.337671 cm Since we only have 3 significant digits, the answer rounds to 10.3 cm</span>
8 0
2 years ago
A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
2 years ago
Which of the following is not a factor in whether a reaction will spontaneously occur? A. Entropy change of the system B. Enthal
Delvig [45]

Answer:

D

Explanation:

pressure change have nothing to do with the spontaneity.

Entropy change , enthalpy change , temperature have roles in deciding spontaneity.

6 0
2 years ago
A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a
o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet m=5 gm

mass of wood block M=1 kg

Length of string L=2 m

Center of mass rises to an height of 0.38 cm

initial velocity of bullet u=450 m/s

let v_1 and v_2 be the velocity of bullet and block after collision

Conserving momentum

mu=mv_1+Mv_2 -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

\frac{Mv_2^2}{2}=Mgh_{cm}

v_2=\sqrt{2gh_{cm}}

v_2=\sqrt{2\times 9.8\times 0.38}

v_2=0.272 m/s

substitute the value of v_2 in equation 1

5\times 450=5\times v_1+1000\times 0.272

v_1=395.6 m/s

4 0
2 years ago
4. Dr. Copus is in charge of the cognition department at the University of Wisconsin-Madison. A new drug named Mem-Reen has beco
MrMuchimi

Answer:

See the answer below

Explanation:

<u>Independent variable</u>: Type of drug (Mem-Reen or placebo)

<u>Dependent variable</u>: memories

<u>Experimental group</u>: The group that was given Mem-Reen

<u>Control group</u>: The group that was given placebo

<u>Constants</u>: Food, hours of sleep, memory test procedures.

The independent variable is an input variable that produces effects on the dependent variable. As the variable is changed, it produces different effects on the dependent variable.

The dependent variable is the actual variable that is measured during an experiment. It is the main purpose of setting-up of an experiment.

The experimental group is also referred to as the treatment group while the control group is the group that does not receive treatment at all or they receive fake treatment/placebo.

Constants are unchanging variables included in experiments. They remain unchanged both in the treatment and the control group, otherwise, the outcome of the experiment will be unreliable.

5 0
2 years ago
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