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MatroZZZ [7]
2 years ago
13

1. Write the advantages and disadvantages of connecting lamps (a) in series and (b) in parallel. A3 2. a) 0.2A Draw Fig 5.30 int

o your exercise book and name the way in which the lamps are connected and answer questions b and c A₂ b) If A3 read 0.2A, what do the other two ammeters read? A1 c) If one of the lamps is removed explain what would happen to the other lamp. Fig 5.30 Reading of ammeter in series circuit​

Physics
1 answer:
spin [16.1K]2 years ago
7 0

Answer:

parallel connection:Advandages: 1. Every unit that is connected in a parallel circuit gets equal amount of voltage.2. It becomes easy to connect or disconnect a new element without affecting the working of other elements.3. If any fault happened to the circuit, then also the current is able to pass through the circuit through different paths.Disadvantages: 1. It requires the use of lot of wires.2. We cannot increase or multiply the voltage in a parallel circuit.3. Parallel connection fails at the time when it is required to pass exactly same amount of current through the units.series connection:Advantages: 1. Series circuits do not overheat easily. This makes them very useful in the case of something that might be around a potentially flammable source, like dry plants or cloth.2. Series circuits are easy to learn and to make. Their simple design is easy to understand, and this means that it’s simple to conduct repairs .3. we can add more power devices, they have a higher output in terms of voltage .4. The current that flows in a series circuit has to flow through every component in the circuit. Therefore, all of the components in a series connection carry the same current.Disadvantages: 1.If one point breaks in the series circuit,the total circuit will break.2. As the number of components in a circuit increases ,greater will be the circuit resistance.

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If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Pachacha [2.7K]

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

4 0
2 years ago
Read 2 more answers
A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang
erma4kov [3.2K]

Given:

initial angular speed, \omega _{i} = 21.5 rad/s

final angular speed, \omega _{f} = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

\alpha = \frac{\omega_{f} - \omega _{i}}{t}

Now, putting the given values in the above formula:

\alpha = \frac{28.0 - 21.5}{3.50}

\alpha = 1.86 m/s^{2}

Therefore, angular acceleration is:

\alpha = 1.86 m/s^{2}

5 0
3 years ago
Identify the name of the most common substance found in the Ocean​
katen-ka-za [31]

Answer:

Explanation:

Salt

3 0
3 years ago
A reaction mixture in a 3.67 L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2, At equilibrium, the f
Alexxandr [17]

Answer:

13 530 482

Explanation:

                            H2    +          I2     ------>      2HI

start (mol)             0.3785         0.3818                   0

change (mol)       -0.3534        -0.3534            +0.7067

equilibrium (mol)  0.0251         0.0284             0.7067

concentra (mol/L) 0.0068        0.0077              0.1926

K_{c} = \frac{0.1926^{2}}{0.0068^{2}*0.0077^{2} } = 13530482

7 0
3 years ago
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