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vitfil [10]
3 years ago
8

Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur

rent of 4.36 A .Is the force per meter exerted on the 4.36-A wire greater than, less than, or the same as the force per meter exerted on the 2.79-A wire
Physics
1 answer:
Dafna1 [17]3 years ago
3 0

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 is the vacuum permeability

I_1 is the current in the 1st wire

I_2 is the current in the 2nd wire

r is the separation between the wires

In this problem

I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m

Substituting, we find the force per unit length on the two wires:

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

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When water freezes, its volume increases by 9.05% (that is, equation). What force per unit area is water capable of exerting on
PIT_PIT [208]

Answer:

1.991 × 10^(8) N/m²

Explanation:

We are told that its volume increases by 9.05%.

Thus; (ΔV/V_o) = 9.05% = 0.0905

To find the force per unit area which is also pressure, we will use bulk modulus formula;

B = Δp(V_o/ΔV)

Making Δp the subject gives;

Δp = B(ΔV/V_o)

Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²

Thus;

Δp = 2.2 × 10^(9)[0.0905]

Δp = 1.991 × 10^(8) N/m²

3 0
3 years ago
1.)Which is the correctly balanced equation?
iVinArrow [24]
A. The correctly balanced equation is that in which the number of atoms of a certain element at the left-hand side is similar to that in the right hand side or the reactant side and product side, respectively. From the given equation, the answer would be,

       C. Cl2 + 2NaI --> 2NaCl + I2


B. In the given chemical reaction above, heat is emitted such that it appears in the product side of the equation. Hence, this is an example of a combustion reaction. 

C. Similar with the reasoning in letter A, the answer to this item is,
          B. 2H2 + O2 --> 2H2O
8 0
3 years ago
A bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away
Varvara68 [4.7K]

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 ×  t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between  when the bat emits the sound and when it hears the echo is 0.05 s.

<h3>How does a bat know how far away something is?</h3>

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

#SPJ4

Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

8 0
2 years ago
Explain why the amplitude of the wave did not change when you increased the frequency of the stimulation. how well did the resul
barxatty [35]
The amplitude did not change when the recurrence was expanded on the grounds that the long headstrong time of the heart forestalls adjustment. It is the most extreme removal or separation moved by a point on a vibrating body or wave measured from its balance position. It is equivalent to the one-a large portion of the length of the vibration way.
5 0
3 years ago
A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

4 0
3 years ago
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