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babunello [35]
4 years ago
10

The electric field in a 3.0mm×3.0mm square aluminum wire is 1.0×10−2 V/m . What is the current in the wire?

Physics
1 answer:
Gekata [30.6K]4 years ago
4 0
So here is my answer. Given that the electric field in <span>a 3.0mm×3.0mm square aluminum wire is 1.0×10−2 V/m, this is how we find the current in the wire.
</span><span>First, we take the distance as 3mm or 0.003m. Using the formula E=V/d, where d is distance, v voltage and E electric field strength, we make V the subject, being V=Ed or 2.2*10^-2*0.003=6.6*10^-5V
</span>Hope this answers your question.
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Uranium (atomic number 92)
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4 years ago
05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86
likoan [24]

Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

Time required to complete one lap around a perfectly circular track is 86 seconds

We need to find the car's velocity. Velocity is equal to,

v=d/t

On circular path,

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 1835}{86}\\\\v=134.06\ m/s

So, car's velocity is 134.06 m/s.

7 0
4 years ago
Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq
Ulleksa [173]

The charge on each the balloon is 100nC or 1.2 × 10^-7 C

Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

The charge on each balloon can be found from

F_{e} = \frac{k_{e}q^{2} }{r^{2}} \\q = \sqrt{\frac{k_{e}q^{2} }{r^{2}}} \\\\q = \sqrt{\frac{(2\times10^{-3N}(0.25m)^{2}}{8.99\times10^{9}N\cdotm^{2}/C^{2}}\\

q = 1.2\times10^{-7}C or 100nC

An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge. The matter is positively charged if it contains more protons than electrons and negatively charged if it contains more electrons than protons.

Learn more about the charge here:

brainly.com/question/14713274

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3 0
1 year ago
A car and a lorry are about to collide. When they collide the two vehicles become tightly locked together. The lorry is going at
BartSMP [9]

Answer:

The speed of the vehicles immediately after the collision is 5.84 m/s.

Explanation:

The speed of the vehicles after the collision can be found by conservation of linear momentum:

p_{i} = p_{f}

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Where:

m₁: is the mass of the car = 0.5 ton = 500 kg

m₂: is the mass of the lorry = 9.5 ton = 9500 kg

v_{1_{i}}: is the initial speed of the car = 40 km/h = 11.11 m/s

v_{2_{i}}: is the initial speed of the lorry = 20 km/h = 5.56 m/s

v_{1_{f}}: is the final speed of the car =?

v_{2_{f}}: is the final speed of the lorry =?    

Since the two vehicles become tightly locked together after the collision v_{1_{f}} = v_{2_{f}}:

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = v(m_{1} + m_{2})

v = \frac{m_{1}v_{1_{i}} + m_{2}v_{2_{i}}}{m_{1} + m_{2}} = \frac{500 kg*11.11 m/s + 9500 kg*5.56 m/s}{500 kg + 9500 kg} = 5.84 m/s

Therefore, the speed of the vehicles immediately after the collision is 5.84 m/s.

I hope it helps you!  

8 0
3 years ago
A lightening bolt may carry a current of 11800 A for a short period of time. The permeability of free space is 1.25664 × 10−6 T
Deffense [45]

Answer:

The resulting magnetic field is 5.021 x 10⁻⁵ T

Explanation:

Given;

current in the lightening bolt, I = 11800 A

distance from the bolt, r = 47 m

permeability of free space, μ₀ = 1.25664 × 10⁻⁶ T· m/A

Assume lightening bolt as long straight conductor, then the resulting magnetic field will be calculated as follows;

B = \frac{\mu_oI}{2 \pi r}

where;

B is the resulting magnetic field

I is the current in the bolt

r is the distance from the bolt

Substitute the given values and calculate B

B = \frac{\mu_oI}{2 \pi r} \\\\B = \frac{1.25664 *10^{-6}*11800}{2 \pi (47)} \\\\B = 5.021 *10^{-5} \ T

Thus, the resulting magnetic field is 5.021 x 10⁻⁵ T

4 0
3 years ago
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