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2 years ago

PLZ HELP ME Which are not correct statements?

1 answer:

8 0

<span>A. Boyle's law only works when the pressure is constant.
</span><span>D. Charles's law relates volume and pressure.

Hope this helps!</span>

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Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire fe

Aleksandr [31]

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

$F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}$

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

$I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A$

Therefore, the bottom current is 12.8 A to the right.

3 0

2 years ago

Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep

Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = $\frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}$

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 - 0:00 = 5

D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 - 0:00 = 6

D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 - 0:00 = 12

D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0

2 years ago

Is it possible to apply the same amount of force and do different amounts of work?

Tamiku [17]

Yes, with simple machines

5 0

3 years ago

Why do nations should establish a set of rules and principles for responsible lunar/moon explortions? ASAP pleaseeee :(

Marianna [84]

So that we do not contaminate it with microorganisms or garbage or other human stuff.

5 0

2 years ago

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit

Stels [109]

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, $P=2.2\times 10^5\ Pa$

Area of each tire, $A=0.023\ m^2$

Area of 4 tires, $A=0.092\ m^2$

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

$P=\dfrac{F}{A}$

$F=P\times A$

$F=2.2\times 10^5\times 0.092$

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

7 0

3 years ago