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If you plan to exercise every day what is something you can do to make it more convenient so that you exercise more regularly?

jolli1 [7]

Answer: exercise doing something you like. for example if you like dancing incorporate dancing in your workouts so you will be more likely to do it regularly

Explanation:

4 0

3 years ago

When a tuning fork vibrates over an open pipe and the air in the pipe starts to vibrate, the vibrations in the tube are caused b

worty [1.4K]

Answer: Resonance!!!

5 0

3 years ago

Design an inverting amplifier that has a gain of -47 (this gain is negative). Pick resistor values that you have in the lab kit.

Nostrana [21]

Answer:

R2= 470 ohms R1= 10ohms. Schematic is attached

Explanation:

Av=-(R2/R1)

-47=-(470/10)

-47=-47

7 0

3 years ago

Why is the efficiency of a machine always less than 100 percent? The work input is too small. It cannot have an IMA greater than

kozerog [31]

Answer:

Some work input is lost to friction

Explanation:

The efficiency of a machine is defined as:

$\eta = \frac{W_{out}}{W_{in}}$ (1)

where

$W_{out}$ is the work output

$W_{in}$ is the work input

Due to the law of conservation of energy, the work output can never be larger than the work input (because energy cannot be created). Moreover, in real machines part of the work input is lost due to the presence of frictions: as a result, part of the energy in input is converted into thermal energy or other forms of energy, and so the work output is smaller than the work input, and so the ratio (1) becomes less than 1, and so the efficiency is less than 100%.

3 0

3 years ago

Read 2 more answers

when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo

Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

$F = k\frac{|q_1|.|q_2|}{r^2}$

Where,

k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

$\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1$ ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

$q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}$

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

$\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6$ .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

$-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123$

$q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\$

6 0

4 years ago