Which is the answer

A 50-V de voltage source was connected in series with a resistor and capacitor. Calculate the current in A to two significant fi

Naddik [55]

Answer:

Current through circuit will be $0.2706\times 10^6A$

Explanation:

We have given source voltage v = 50 volt

Resistance $R=25Mohm=25\times 10^6ohm$

Capacitance $C=0.1\mu F=0.1\times 10^{-6}F$

Time t = 5 sec

Time constant of RC circuit is given by $\tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec$

We know that voltage across capacitor is given by $v_c=v_s(1-e^{\frac{-t}{\tau }})$

$v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v$

So current will be $=\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A$

So current through circuit will be $0.2706\times 10^6A$

4 0

4 years ago

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is:

$a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)$

$a_{t} = -1.075\, \frac{m}{s^{2}}$

The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) If we know that $\omega_{o} = 104.720\,\frac{rad}{s}$ and $\omega = 26.180\,\frac{rad}{s}$ , then the time needed is:

$26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}$

$5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}$

$t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5}$

$t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5} }$

$t \approx 3.963\,s$

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

8 0

3 years ago

A soccer player attempting to steal the ball from an opponent was extending her knee at 50 deg/s when her foot struck the oppone

denpristay [2]

Answer:

$\alpha = -4.36 rad/s^2$

Explanation:

Initial angular speed of the player is given as

$\omega = 50 deg/s$

here we know that

$1 deg = \frac{\pi}{180} rad$

now we know that

$\omega = 50\frac{\pi}{180}$

$\omega = 0.87 rad/s$

now its speed comes to zero in 0.2 s

so angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0 - 0.87 rad/s}{0.2 s}$

$\alpha = -4.36 rad/s^2$

7 0

3 years ago

When we see a region of a planet that is not as heavily cratered as other regions, it means that

balandron [24]

Answer:

When we see a region of a planet that is not as heavily cratered as other regions, it means that the surface of that region is younger.

Explanation:

There are a number of processes that affect planetary surfaces like erosion, volcanism, plate tectonics and the amount of craters by impact of meteorites.

The density of craters in a given region suggests the age of the surface. If the surface is saturated by craters then is likely very old because a long time should have passed for the planet to be hit by so many meteorites. If there are very few craters this may indicate that the surface is quite young, because the other processes described above wouldn't have had enough time to erase them. In any case we need to study the change of the cratering rate over time to determine the age of a given surface.

7 0

3 years ago

I need the answer with steps

Elis [28]

Answer: What Question

Explanation:

6 0

3 years ago