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ziro4ka [17]
3 years ago
8

How does an antacid table relieve an upset stomach

Physics
1 answer:
VashaNatasha [74]3 years ago
6 0
The antacid reacts with the stomach acid to relieve the upset stomach
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Which object has a volume greater than the volume of a pencil? *
Bad White [126]

Answer:

A.

Explanation:

3 0
3 years ago
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A ball is projected at an immovable wall with a speed vi and bounces back the wall in such a manner that it only has 1/3 of its
sergey [27]

The fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

The given parameters;

  • <em>initial speed of the ball, = vi</em>
  • <em>final momentum of the ball, Pf = ¹/₃Pi</em>

The initial and final momentum of the ball is calculated as;

P_i = m_ivi

P_f = m_fv_f = \frac{1}{3} m_iv_i

The initial and final kinetic energy of the ball is calculated as;

K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i

The change in the kinetic energy is calculated as;

\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i

Thus, the fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

Learn more here:brainly.com/question/18566218

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3 years ago
4. Can an object's displacement be negative? What does the sign indicate about the<br> displacement?
faltersainse [42]
Displacement can be positive OR negative. The sign implies the direction of the displacement. Negative means that is is moving to the left (sometimes called West) or down (sometimes called South); positive means that is moving to the right (sometimes called East) or up (sometimes called North).
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Can the number of protons in an element ever change?
adelina 88 [10]
The number of protons in any element can never change.
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Kindly Don't Spàm!<br>Thank uh !:)<br><img src="https://tex.z-dn.net/?f=%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20" id="Te
olga55 [171]

.

\:  \underline{ \boxed{  { \color{gray}\frak{\huge \: Answer : }}}}

<h3> Collector Current at Saturation :</h3>

  • \sf \large {I_{C(SAT)} = \frac{ V_{CC} }{R_C} }

\:  \:

  • \sf \large {I_{C(SAT)} = \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bold{ \bf \large {I_{C(SAT)} =5.45mA }}

\:  \:

________________________________

<h3> Value Of Cut - off Voltage : </h3>

  • \sf \large \: V_{CS(cut-off)} = V_{CC}

\:  \:

Therefore ,

\:  \:

  • \bf \large \: V_{CS(cut-off)} = \: 12v

\:  \:

________________________________

{ \bf\large \: Base \:  Current , I_B =  \frac{V_{CC}}{R_C} }

  • \sf \large \: I_B ={ \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bf \large \: I_B = 50μA

\:  \:

_______________________________

<h3>Collector Current ,</h3>

  • \sf \large \: I_C = β * I_B

\:  \:

  • \sf \large \: I_C = 50 * 50 * 10^{-6}

\:  \:

  • \bf \large \: I_C = 2.5mA

\:  \:

________________________________

<h3> Collector to emitter Voltage : </h3>

  • \sf \large \: V_{CE} = V_{CC} - ( I_C * R_C )

\:  \:  \:

  • \sf \large \: V_{CE} = 12 - ( 2.5 * 10-³ * 2.2 * 10³ )

\:  \:

  • \bf \large \: V_{CE} = 6.5v

________________________________

<h3>Q - point are :</h3>

  • \sf \large \: I_{CEQ} = 2.5mA

\:  \:

  • \sf \large \: V_{CEQ} = 6.5v

________________________________

<h3>Q - point located on the DC load line as shown in fig ~</h3><h3 /><h3 /><h3 /><h3 />

________________________________

Hope Helps!:)

\:  \:  \:  \:

5 0
2 years ago
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