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valina [46]
3 years ago
11

In Florida, once you have had your learner's license for _________________ without any traffic convictions, you will receive an

intermediate license.
Physics
1 answer:
bogdanovich [222]3 years ago
3 0

Answer:

One year

Explanation:

In Florida, once you have had your learner's license for 1 year without any traffic convictions, you will receive an intermediate license.

The rule for Florida is that if a driver drives safely without any conviction during that year of his trial, he will receive an intermediate license. This means that he is now completely eligible and safe to drive around Florida's roads.

I hope the answer is helpful.

Thanks for asking.

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Two iron rods are placed horizontally, touching end-to-end. One end of one rod is heated. After some time, the other rod also fe
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The answer is D) <span>The molecular movement in the first rod transferred energy to the molecules in the second.</span>
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4 years ago
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A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of
OLEGan [10]

Answer:

c)F_{net} = 0.640 kN

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

F_{net} = ma

here we know

m = 80 kg

for circular motion acceleration is given as

a_c = \frac{v^2}{R}

a_c = \frac{40^2}{200} = 8 m/s^2

now we have

F_{net} = 80 \times 8

F_{net} = 640 N

F_{net} = 0.640 kN

7 0
3 years ago
A cell phone weighing 80 grams is flying through the air at 15 m/s. What is its kinetic energy
Contact [7]

Answer:

The kinetic energy of the cell phone is 9J

Explanation:

The kinetic energy is the energy possessed by a body by virtue of motion.

The kinetic energy is expressed as

KE= 1/2m(v)²

Given data

Mass of cell phone m= 80g--to kg=80/1000= 0.08kg

Velocity of cell phone v= 15m/s

Substituting our given data we have

KE= 1/2*0.08(15)²

KE= (0.08*225)/2

KE=18/2

KE= 9J

8 0
3 years ago
The electron drift speed in a 1.2-mm-diameter gold wire is 5.00 10-5 m/s. how long does it take 1 mole of electrons to flow thro
Vlada [557]
Twenty four minutes per second

7 0
3 years ago
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
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The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
3 years ago
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