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Inga [223]
3 years ago
7

A sharp raises a note by

Physics
2 answers:
vivado [14]3 years ago
8 0

The answer should be B. A half step

-Dominant- [34]3 years ago
6 0
The flat lowers a note by a half step while the sharp raises a note by a half step. So your answer is B. A half step
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Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the
GarryVolchara [31]

Explanation:

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).

For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:

P^2=kA^3

k is constant of proportionality

It is required to solve the above equation for k

k=\dfrac{P^2}{A^3}

8 0
3 years ago
The term _______ is used to indicate the frequency level of a sound.
aksik [14]
The answer is D, pitch

6 0
3 years ago
Read 2 more answers
Pedro is planning to model how changes in weather affect evaporation from lakes for his first experiment he wants to test how hu
Valentin [98]

Answer:

C. volume of water and temperature

Explanation:

a p e x

6 0
3 years ago
11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of
Marat540 [252]

a =  \frac{v - u}{t}

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

\frac{10 - 0}{3}  = 3.333333....

approximately 3.33 m/s^2

the acceleration of the car is

\frac{40 - 0 }{8}  = 5.0

5.0 m/s^2

5.0 > 3.33 \\ so \:  the \: answer  \: is \: no

6 0
3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
3 years ago
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