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erma4kov [3.2K]
3 years ago
12

The drawing shows a wire composed of three segments, AB, BC, and CD. There is a current of I = 2.0 A in the wire. There is also

a magnetic field B (magnitude = 0.25 T) that is the same everywhere and points in the direction of +z axis. The lengths of the wire segments are LAB = 1.0 m, LBC = 0.7 m, and LCD = 0.7 m. Find the magnitude of the force that acts on each segment.
Physics
1 answer:
alexdok [17]3 years ago
7 0

Answer:

The magnitude of the magnetic force acting on the wire is zero, because the magnetic field is parallel to the wire.

In fact, the magnetic force exerted by the magnetic field on the wire is

where I is the current in the wire, L the length of the wire, B the magnetic field intensity and  the angle between the direction of B and the wire. In our problem, B and the wire are parallel, so the angle is  and so , therefore the magnetic force is zero: F=0.

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An object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34
Travka [436]

The initial position of the object was found to be 134.09 m.

<u>Explanation:</u>

As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

Initial position of the object = Final position-Displacement

Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.

Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

4 0
3 years ago
Is xenon a pure substance​
forsale [732]

\large\huge\green{\sf{Yes}}

6 0
2 years ago
Read 2 more answers
It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W
Anvisha [2.4K]

Answer:

962291.57928 m²

Explanation:

P_r = Pressure = 2\dfrac{I}{c}  (full reflection)

I = Intensity = \dfrac{P}{A}=\dfrac{P}{4\pi r^2}

P = Power = 3.9\times 10^{26}\ W

c = Speed of light = 3\times 10^8\ m/s

M = Mass of Sun = 1.99\times 10^{30}\ kg

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A

This force will balance the gravitational force as stated in the question

\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2

The area of the must be 962291.57928 m²

3 0
3 years ago
At 19 ºC a volume of air takes us 9.0 mL in an expandable container. What is its volume after it is heated up to 50. ºC?
Katena32 [7]
According to Charles law, we know, at constant pressure, volume is directly proportional to temperature. 
So, <span>V/T = constant 
</span>
V₁/t₁ = V₂/t₂
V₁t₂ = V₂t₁

Here, we have: V₁ = 9 mL
V₂ = ?
T₂ = 50+272 = 323 K
T₁ = 19+273 = 292 K

Substitute their values into the expression:
9 × 323 = V₂ × 292
V₂ = 2907 / 292
V₂ = 9.95
After rounding-off to unit place value, it would be equal to 10 mL

So, In short Option C would be your correct answer.

Hope this helps!
5 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
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