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ki77a [65]
3 years ago
11

2. Why do you need eggs to make manynaise? (yo

Physics
1 answer:
anastassius [24]3 years ago
6 0

Answer:

to make it fluffy

Explanation:

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An RC circuit consists of a resistor with resistance 1.0 kΩ, a 120-V battery, and two capacitors, C1 and C2, with capacitances o
Serhud [2]

Answer:

Q_t= 8.3 * 10^3 C

Explanation:

From the question we are told that:

Resistor R=1000ohms

Voltage v=120_V

Capacitance of c_1 c_1=20 \mu F

Capacitance of c_2 c_2=60 \mu F

Time t=0

Generally the equation for charges is mathematically given by

For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\

ForC_2\\Charge on C_2 = 60*120 =7200 μC =  7.2 x 10^-3\\Charge on C_2 =  7.2 x 10^-3

Generally the equation for voltage across capacitors is mathematically given by

V_c(t)=V(1-e^{-t/RC})

C=C_1+C_2=80 \mu f\\t=2RC=>160000s

V_c(t)=120(1-e^{-(160000)/1000*(80)})

V_c(t)=103.7598

Generally the equation for charges is mathematically given by

Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\

Generally the equation for total charges Q_t is mathematically given by

Q_t=Q1(t)+Q2(t)

Q_t= 8.3 * 10^3 C

5 0
3 years ago
What are some important factors to consider when choosing a warm-up before your workout?
TEA [102]

<u>Answer:</u>

Prior to exercise, a proper warm-up of 10-15 minutes is extremely important to avoid injuries.

  1. Don't go too hard in the beginning and boost your activity level slowly. A good indication of a proper warm-up is that you feel sweat on your body parts.
  2. Don't overstretch right in the beginning as it can cause sore in your muscles and joints or stress fractures.
  3. Take a break if you feel sick or fatigues and use other drinks along with water to replace electrolytes and body fluids.  
6 0
4 years ago
Read 2 more answers
Potassium and chlorine are most likely to form what type of bond? ionic covalent metallic polar
fiasKO [112]

Answer:

The correct answer is the IONIC bond.

Explanation:

4 0
4 years ago
Read 2 more answers
For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par
Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

  • <em>initial temperature of metals, =  </em>T_m<em />
  • <em>initial temperature of water, = </em>T_i<em> </em>
  • <em>specific heat capacity of copper, </em>C_p<em> = 0.385 J/g.K</em>
  • <em>specific heat capacity of aluminum, </em>C_A = 0.9 J/g.K
  • <em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w )  = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w)  = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0
3 years ago
Find the orbital period (in years) of an asteroid whose average distance from the sun is 27
dmitriy555 [2]
Using Kepler's third law which is defined as the square of the average distance is directly proportional to the cube of the period. It is expressed as P^2 = a^3, Given that the a = average distance is given, the period would be much easier to compute. P = sqrt(27^3) = 140
7 0
3 years ago
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