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Elanso [62]
1 year ago
5

What is the direction of the magnetic field b⃗ net at point a? Recall that the currents in the two wires have equal magnitudes.

Physics
1 answer:
andrew11 [14]1 year ago
6 0

Answer:

Explanation:

The direction of a magnetic field indicates where the magnetic inluence on the electric charges are directed to.

From the given  question, we are to determine the direction of the magnetic field bnet at a point A.

Also, having the notion that  the currents in the two wires have equal magnitudes, Then:

\bar{B_{net}} = \bar{B_1} + \bar{B_2}

\bar{B_{net}} = \frac{\mu_oI}{2 \pi r } \bar {k}+ \frac{\mu_oI}{2 \pi r } \bar {k}

\bar{B_{net}} = \frac{2 \mu_oI}{2 \pi r } \bar {k} \ out

Thus; \bar{B_{net}} points out of the screen at A.

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Un electrón en un tubo de rayos catódicos acelera desde el reposo con una aceleración constante de 5.33x10¹²m/s² durante 0.150μs
Andre45 [30]
Can you translate that in English ? I'll try to help you out with that..
3 0
2 years ago
A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil
adelina 88 [10]
<span>14 m/s Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration. 1. How much energy is stored in the spring? 2. How fast will the dart travel with that amount of energy. As for the energy stored, that's a simple matter of multiplication. So: 20 N * 0.05 m = 1 Nm = 1 J For the second part, the energy of a moving object is expressed as KE = 0.5 mv^2 where KE = Kinetic energy m = mass v = velocity Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So KE = 0.5 mv^2 1 J = 0.5 0.010 kg * v^2 1 kg*m^2/s^2 = 0.005 kg * v^2 200 m^2/s^2 = v^2 14.14213562 m/s = v So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>
6 0
2 years ago
Read 2 more answers
The curved section of a horizontal highway is a circular unbanked arc of radius 740m. If the coefficient of static friction betw
fiasKO [112]
Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²
v \:  =  \:  \sqrt{gr \tan( \alpha ) }
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s
6 0
2 years ago
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To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch
sergij07 [2.7K]

Answer:

man upward acceleration is 0.14m/s^2

Explanation:

given data:

mass of man = 72 kg

downward force = 360 N

The mass of man of weight 72 kg is hang from two sections of rope, one section pf rope ties around man waist and other section is ties in man hands. when he pulls down the rope  with 360 N force then each section of  rope pulls with 360 N

we know that

Weight= mass × gravity= 72kg × 9.8 = 705.6N

Force = mass× acceleration

Force= -705.6 + (2 × 358) = 10.4 N

acceleration = \frac{10.4}{72} = 0.14m/s^2

4 0
1 year ago
The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an
Troyanec [42]

Answer:

3400 Hz

Explanation:

We know that

1 cm = 0.01 m

L = Length of the human ear canal = 2.5 cm = 0.025 m

V = Speed of sound = 340 ms⁻¹

f = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

f = \frac{(2n - 1)V}{4L}

for first resonant frequency, we have n = 1

Inserting the values

f = \frac{(2(1) - 1) 340}{4(0.025)}

f = \frac{340}{4(0.025)}

f = 3400 Hz

4 0
2 years ago
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