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3 years ago

The variable that you change during an experiment is called what type of variable?

Physics

2 answers:

elena-s [515]3 years ago

7 0

The independent variable

Norma-Jean [14]3 years ago

5 0

Independent is the affect of a dependent variable
control keeps everything the same

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The small metal cylinder has a mass of 0.20 kgkg, the coefficient of static friction between the cylinder and the turntable is 0

deff fn [24]

Answer:

velocity of the metal cylinder = 0.343 m/s

Explanation:

Force putting the metal cylinder is given by

F = mv²/r

But this force will balance the frictional force between the metal cylinder and the turntable

The frictional force is given by

μN = μ × mg = 0.08 × 0.2 × 9.81 = 0.15696 N

r = 0.15 m, m = 0.2 kg,

F = mv²/r = 0.2 v²/(0.15) = 1.3333 v²

1.3333 v² = 0.15696

v² = 0.117

v = 0.343 m/s

8 0

3 years ago

A proton is released in a uniform electric field, and it experiences an electric force of 5 X 10^-10 N toward the South. What is

Ymorist [56]

Answer:

Electric field on proton

$E=3.12\times 10^9\ N/C$

Explanation:

Given that

$Force,F=5\times 10^{-10}\ N$

We know that

Charge on proton

$q=1.6\times 10^{-19}\ C$

We know that

Force = Electric field x Charge

F= E x q

$E=\dfrac{F}{q}\ N/C$

$E=\dfrac{5\times 10^{-10}}{1.6\times 10^{-19}}\ N/C$

$E=3.12\times 10^9\ N/C$

Electric field on proton

$E=3.12\times 10^9\ N/C$

4 0

3 years ago

When the moon is waxing, it's lighted part appears

kvv77 [185]

To increase it's size

Waxing is the opposite of waning, which is to decrease.

7 0

3 years ago

Read 2 more answers

If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re

-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = $\frac{1}{2}kx^2$ ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = $\frac{1}{2}k\times0.021^2$

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = $\frac{1}{2}\times13605.44\times (0.062)^2$

U = 26.149 J

Also,

Work done = change in energy

W = 26.149 - 3.0 = 23.149 J

4 0

2 years ago

A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r

Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l = 4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

$F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s$

Let $\omega$ is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

$v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s$

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0

3 years ago