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1 year ago

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The oscillating current in an electrical circuit is as follows, where I is measured in amperes and t is measured in seconds. I =

7 sin(60πt) + cos(120πt). Find the average current for each time interval. (Round your answers to three decimal places.) (a) 0 ≤ t ≤ 1 60 amps

1 answer:

Vesna [10]1 year ago

3 0

Answer:

Explanation:

Given

$I=7\sin (60\pi t)+\cos (120\pi t)$

$=\frac{1}{\frac{1}{60}}\times \int_{0}^{\frac{1}{60}}7\sin (60\pi t)dt+\int_{0}^{\frac{1}{60}}\cos (120\pi t)dt$

$=60\left [ \frac{-7}{60\pi }\cos (60\pi t)+\frac{1}{120\pi }\sin (120\pi t)\right ]_{0}^{\frac{1}{60}}$

$=60\left [ \frac{-7}{60\pi }\left ( \cos \pi-\cos 0\right )+\frac{1}{120\pi }\left ( \sin (2\pi )-\sin (0)\right )\right ]$

$=\frac{-7}{\pi }\left ( -1-1\right )+\frac{1}{2\pi }\left ( 0-0\right )$

$=\frac{14}{\pi }$

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A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle

s344n2d4d5 [400]

Answer:

0.24 kgm²

Explanation:

$L$ = length of the bat = 81.3 cm = 0.813 m

$m$ = mass of the bat = 0.96 kg

$d$ = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

$T$ = Period of oscillation = 1.35 sec

$I$ = moment of inertia of the bat

Period of oscillation is given as

$T = 2\pi \sqrt{\frac{I}{mgd}}$

$1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}$

$I$ = 0.24 kgm²

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2 years ago

A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t

Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature $T_1=323K$

Volume $V_1=V\ and\ V_2=\frac{V}{2}$

As there is no heat transfer so this is an adiabatic process

For and adiabatic process $TV^{\gamma -1}=constant$

Here $\gamma =1.4$

So $T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}$

$T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1$

$T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K$

4 0

2 years ago

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha

devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_{max}=\frac{hc}{\lambda}-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{495*10^{-9}m}-0.5eV\\W=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0

2 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

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1 year ago

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed

olga55 [171]

Initially, the energies are:

$U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
=K_{i}=\frac{1}{2} m v_{0}^{2}$

At final point, the energies are:

$U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0$

Using conservation law of energy,

$-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}$

The equation is further simplified as:

$r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\
h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\
&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\
& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}$

7 0

1 year ago