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yarga [219]
3 years ago
15

For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

rcuit? the current leads the generator EMF by 0.982 radians, what are IT 3 20 the resistance value of the resistor, R?
Physics
1 answer:
MAXImum [283]3 years ago
5 0

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

z = \frac{V_o}{i_o}

now we will have

z = \frac{125}{3.20} = 39.06 ohm

Part b)

Now we also know that

\frac{R}{z} = cos\theta

\theta = 0.982 rad = 56.3 degree

now we have

\frac{R}{39.06} = cos56.3

R = 21.7 ohm

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A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr
Zinaida [17]

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

m'=9.00 \ g

or,

    =0.009 \ kg

mass of wooden block,

m=1.20 \ kg

speed,

s=0.390 \ m

Coefficient of kinetic friction,

\mu=0.20

As we know,

The Kinematic equation is:

⇒  v^2=u^2+2as

then,

Initial velocity will be:

⇒  u=v^2-2as

        =v^2-2 \mu gs

On substituting the given values, we get

⇒  u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}

       =\sqrt{-1.5288}

       =1.23 \ m/s

As we know,

The conservation of momentum is:

⇒  mu=m'u'

or,

⇒ Initial speed, u'=\frac{mu}{m'}

On substituting the values, we get

⇒                            =\frac{1.20\times 1.23}{0.009}

⇒                            =\frac{1.476}{0.009}

⇒                            =164 \ m/s                              

3 0
2 years ago
5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t
Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

5 0
3 years ago
The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

5 0
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Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c
grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg              

Force applied by the motor, F_A=1000\ N

The static and dynamic friction coefficient is, \mu=0.5

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

F_A-\mu mg=ma

\dfrac{F_A-\mu mg}{m}=a

a=\dfrac{1000-0.5(1000)(9.81)}{1000}

a=-3.905\ m/s^2

So, the acceleration of the car is -3.905\ m/s^2. Hence, this is the required solution.

6 0
2 years ago
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