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Mrrafil [7]
3 years ago
5

When two magnets are brought close together their magnetic

Physics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

You know the saying "Opposites attract" well that is how you can remember that South and North Magnetic Poles connect.  

Explanation:

Hope this helps ya

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A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V
yan [13]
<h2>Answer:</h2>

(a) Attached to the response as Figure 1.

(b) 35.0Ω

(c) Across 5.0Ω = 1.3V

   Across 10.0Ω = 2.6Ω

   Across 20.0Ω = 5.2Ω

<h2>Explanation:</h2>

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃               -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

<em>Substitute these values into equation (i) as follows;</em>

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R              ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = \frac{9.0}{35.0}

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1.<em> 5.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2.<em> 10.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3.<em> 20.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

7 0
3 years ago
Two rams run toward each other. One ram has a mass of 49 kg and runs west
olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram m_1=49\ kg

The velocity of this ram is v_1=-7\ m/s

Mass of the second ram m_2=52\ kg

The velocity of this ram v_2=9\ m/s

They combined after the collision

Conserving the momentum

\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]

Momentum after the collision will be

\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}

Therefore, option (d) is correct

4 0
2 years ago
A Jack Rabbit hops 12.8 meters per second. How long would it take for him to hop 353 m? Round to the nearest tenth place. ​
adoni [48]

Time = Distance/speed

353/12.8 =27.57 seconds

4 0
2 years ago
The current in a stream runs at 5 miles per hour. if a boat can go 15 miles per hour on still water, how fast can the boat go do
arsen [322]
Down stream it would be going 20 mph and up stream 10 mph
3 0
3 years ago
An air conditioner operating at steady state maintains a dwelling at 70oF on a day when the outside temperature is 90oF. If the
torisob [31]
30000 btuh /3413 btuh/kw. = 8.8 kw

8.8 kw/.746 kw/hp = 11.8 hp if COP is 1

11.8/3 hp (COP coefficient of performance) = 3.99 COP

>>>So yes a 3.0 hp compressor with a nominal COP of 4 will handle the 30,000 btuh load.

3.2 to 4.5 is expected COP range for an air cooled heat pump or a/c unit.
8 0
3 years ago
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