Which should not be a part of scientific inquiry?

A cardio kickboxing class challenges 3 things. List them.

bagirrra123 [75]

Answer:

In a cardio kickboxing class, you will learn proper form for the famous jab, hooks,crosses, uppercuts, back kicks, front kicks.

Explanation:

6 0

2 years ago

The diagram shows parts of a wave. A series of waves with an arrow passing through their centers. The highest point of one wave

Pachacha [2.7K]

Answer:

im not sure but i think its c

Explanation:

3 0

3 years ago

Read 2 more answers

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

2 years ago

To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is acc

levacccp [35]

The formula for resonant frequency is:

$f_0=\frac{1}{2\pi\sqrt{LC} }$

Given information:

$f_{0\text{,small}}=500 \text{ kHz}\\f_{0\text{,large}}=1650 \text{ kHz}\\L=3.83\text{ } \mu \text{H}$

Plug in the given values to find one value of capacitance:

$500 \text{ kHz}=\frac{1}{2\pi\sqrt{C(3.83\text{ } \mu \text{H})} }\\C=2.645*10^{-8} \text{ F}=26.45 \text{ nF}$

Plug in the given values to find the other value of capacitance:

$1650 \text{ kHz}=\frac{1}{2\pi\sqrt{C(3.83\text{ } \mu \text{H})} }\\C=2.429*10^{-8} \text{ F}=2.429 \text{ nF}$

This gives a range of 2.429 nF to 26.45 nF.

With significant figures taken into account, the range of capacitance is 2.43 nF to 30 nF.

6 0

2 years ago

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A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increas

Westkost [7]

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

$\Phi = BA Cos\theta$

Where

B= Magnetic Field

A = Area

$\theta =$ Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

$\theta =$ 0 then our expression can be written as

$\Phi = BA$