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3 years ago

Resistance increases with the length of the wire

1 answer:

4 0

If the length of the wire increases, then the amount of resistance will also increase.

1. Take a long piece of wire and cut it 10 pieces. Those pieces should all be different sizes, one should be 5___ (units in meter, cm, inches, etc.), and the next should be 5 ___ (units in meter, cm, inches, etc.) more than the one before.
2. Take one piece of wire and measure the resistance using ___ and record the results in the data table.
3. Repeat the previous step with all the pieces of wire.
4. Compare and contrast the results you have found.

I hope this helps a bit :)

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A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each

nataly862011 [7]

The centripetal acceleration is $13.0 m/s^2$

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference ( $2\pi r$ ) and the period of revolution (T), so it can be rewritten as

$v=\frac{2\pi r}{T}$

Therefore we can rewrite the acceleration as

$a=\frac{4\pi^2 r}{T^2}$

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

$T=\frac{1}{4}s = 0.25 s$

Substituting into the equation, we find the acceleration:

$a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2$

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0

2 years ago

The study of charges in motion and their

vredina [299]

Answer:

The study of charges in motion and their interaction with magnetic fields is known as electromagnetism.

Any material with an unequal number of electrons and or protons could generally be termed as ion.

Explanation:

i hope this will help you :)

4 0

2 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

1 year ago

Help please? I dont understand what this means.

Alex Ar [27]

The answer is c for real

5 0

2 years ago

Is it possible to do work on an object without changing the kinetic energy of the object? Now Why?

Murrr4er [49]

Answer:

(a) Yes, it is possible by raising the object to a greater height without acceleration.

Explanation:

The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.

If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.

If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).

Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.

Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".

8 0

2 years ago