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Alina [70]
3 years ago
8

You connect a 250-2 resistor, a 1.20-mH inductor, and a 1.80-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. When

you connect a voltmeter across the resistor in this circuit, the meter reads approximately A) 25.7 mV B) 83.6 V C) 120 V D) 14.2 V E) 84.9 V
Physics
1 answer:
Basile [38]3 years ago
4 0

Answer:

Explanation:

 reactance of inductor = wL = 2 X 3.14 X 60 X 1.2 X 10⁻³ = .45 ohm.

reactance of capacitor = 1/wC = 1/( 2 X 3.14 X 60 X 1.8 X 10⁻⁶ ) = 1474.4

Impedence of the circuit =[ R² + ( I/wC - wL)  ]¹/²  = [250² + ( 1474.4-.45 )]¹/²

Impedence = 1495 ohm.

RMS Voltage = 120/ 1.414 = 84.86 V

current = 84.86 / 1495 = 0.0576

Potential over resistance = 0.0576 x 250 = 14.2 V.

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A loader sack of total mass
vampirchik [111]

Question: A loader sack of total mass

is l000 grams falls down from

the floor of a lorry 200 cm high

Calculate the workdone by the

gravity of the load.​

Answer:

19.6 Joules

Explanation:

Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

W = (1×2×9.8)

W = 19.6 Joules

Hence the work done by gravity on the load is 19.6 Joules

8 0
3 years ago
Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d
dezoksy [38]

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J

Hence, the required work done is 1786.17 J.

6 0
3 years ago
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2.​ We want
lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

\Delta x=v_ot+\dfrac{1}{2}at^2

t is time taken by the ball to hit the ground

v_o is initial speed of the ball

So, the correct option is (A).

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3 years ago
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2. The resistance of a conductor depends on the cross sectional area of the conductor and it's resistivity.

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4.Due to at high temperatures , the alloy donot oxidize. Alloy doesn't melt readily and get deformed.

8 0
3 years ago
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