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2 years ago

Which copper wire would have the lowest resistance?

Physics

2 answers:

Ronch [10]2 years ago

6 0

B. The resistance is directly proportional to length and inversely proportional to cross sectional area

vodomira [7]2 years ago

3 0

What would be the highest resistance ?

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A proton is confined to a space 1 fm wide (about the size of an atomic nucleus). What's the minimum uncertainty in its velocity?

Daniel [21]

Answer:

Minimum uncertainty in velocity of a proton, $\Delta v\ge 3.15\times 10^7\ m/s$

Explanation:

It is given that,

A proton is confined to a space 1 fm wide, $\Delta x=10^{-15}\ m$

We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,

$\Delta p.\Delta x\ge \dfrac{h}{4\pi}$

Since, p = mv

$\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}$

$m \Delta v.\Delta x\ge \dfrac{h}{4\pi}$

$\Delta v\ge \dfrac{h}{4\pi m\Delta x}$

$\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}$

$\Delta v\ge 3.15\times 10^7\ m/s$

So, the minimum uncertainty in its velocity is greater than $3.15\times 10^7\ m/s$ . Hence, this is the required solution.

3 0

3 years ago

When drawing a Bohr model for an element that has 16 electrons, how many electrons would be placed in the third energy level?

lubasha [3.4K]

There would be 6 electrons placed on the third energy level.

5 0

2 years ago

A skateboarder traveling with an initial velocity 9.0 meters per second,

meriva

Answer:

25m/s

Steps:

First, The equation v= u + a * t shows us what we need to find, (the finale velocity).

Second, we substitute the values given:

v= 9m/s + 4m/s2 * 4s

Last, We calculate the values:

Multiply 4m/s2 * 4s = 16m/s

Add 9m/s + 16m/s

Answer: 25m/s

Hope this helps :)

4 0

2 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0

2 years ago

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun

inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

$s=-8t^2+32t$ ...........(1)