All mutations are favorable and increase the ability for the

Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.40 kg mass to the

Aleks [24]

Answer:

Explanation:

mg = kx

k = mg/x

k = 0.40(9.8)/0.030

k = 130.66666...

k ≈ 130 N/m

4 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

An object is raised 2.5m off the ground. That object has a mass of 10kg, how many joules of potential energy is stored in the ob

klio [65]

Answer:

245.25 J

Explanation:

Potential Energy = m g h

= 10 * 9.81 * 2.5 = 245.25 J

4 0

3 years ago

A iron block with a mass of 4.8 kg initially slides over a rough horizontal surface with a speed of 1.2 m/s. Friction slows the

Harlamova29_29 [7]

Answer:

Explanation:

Kinetic energy of block will be converted into heat energy by friction .

Heat energy produced = 1/2 m v²

= .5 x 4.8 x 1.2²

= 3.456 J

85% of energy is converted into heat energy , so heat energy produced

= .85 x 3.456 = 2.9376 J .

If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature

Q = msΔt

specific heat of iron s = 462 J / kg C

Putting the values ,

2.9376 = 4.8 x 462 x Δt

Δt = 13.24 x 10⁻⁴ ⁰C.

4 0

3 years ago

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to

Nostrana [21]

Answer:

a) r=4.24cm d=1 cm

b) $Q=5x10^{-10} C$

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

$V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}$

$V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m$

The distance must be the separation the r distance can be find also using

$C=\frac{Q}{V_{ab}}$

But now don't know the charge these plates can hold yet so

a).

d=0.01m

$C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}$

$A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}$

$A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } } \\r=42.55x^{-3}m$

b).

$C=\frac{Q}{V_{ab}}$

$Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C$

5 0

3 years ago