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zheka24 [161]
3 years ago
9

scientists use critical thinking skills throughout the process of research or information a student noticed that their energy le

vel always seems to decrease when they drink soda but they seem to have more energy when they don't drink soda the student decides to conduct an experiment yo determine if their soda intake really does have an effect on their energy level what critical thinking step has the student just completed in the situation describe above ?
Physics
1 answer:
Ganezh [65]3 years ago
5 0

The critical thinking step has the student just completed in the situation describe when the student concluded that soda intake really does have an effect on their energy level is the analysis. This is because the student observed the effects of drinking soda by his experience.

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For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr
Marina CMI [18]

Answer:

 L= 2 mH

C=1.26\times 10^{-7}\ F            

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

E=\dfrac{1}{2}LI^2

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

10^{-5}=\dfrac{1}{2}\times L\times 0.1^2

L=\dfrac{2\times 10^{-5}}{0.1^2}\ H

L=0.002 H

L= 2 mH

We know that frequency f is given as

2\pi f=\dfrac{1}{\sqrt{LC}}

C=Capacitance  , f=frequency ,L=Inductance

Now by putting the values

2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }

62831.85=\dfrac{1}{\sqrt{0.002\times C}}

\sqrt{0.002\times C=\dfrac{1}{62831.85}

0.002\times C=0.0000159^2

C=\dfrac{0.0000159^2}{0.002}\ F

C=1.26\times 10^{-7}\ F

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

6 0
3 years ago
A spinning disc rotating at 130 rev/min slows and stops 31 s later. how many revolutions did the disc make during this time?
gayaneshka [121]
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
3 0
3 years ago
A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0
3 years ago
A loudspeaker produces a sound wave of frequency 50hz.
viktelen [127]
Hey mate
Here is your answer
Option A)
Explanation:
The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.
Pls mark as brainliest
3 0
3 years ago
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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