All categories

2 years ago

In the diagram, R1 = 40.0 ,

Physics

1 answer:

Nostrana [21]2 years ago

4 0

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

You might be interested in

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

Explanation:

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

3 0

3 years ago

What volume will be occupied by 20g of Copper if the density of Copper is 9.1g/ml?

Eva8 [605]

Answer:

The volume of copper is 2.198 ml

Explanation:

Given;

mass of copper, m = 20 g

density of copper, ρ = 9.1 g/ml

Density is given by;

Density = mass / volume

Volume = mass / density

Volume = (20 g) / (9.1 g/ml)

Volume = 2.198 ml

Therefore, the volume of copper is 2.198 ml

3 0

2 years ago

Sound wave A delivers 2 J of energy in 2 s. Sound wave B delivers 10 J of energy in 5 s. Sound wave C delivers 2 mJ of energy in

Dmitry [639]

Answer:

In the order largest to smallest

Pc = Pb, Pa

Explanation:

Power = Energy expended/time

Pa = 2/2 = 1Watt

Pb = 10/5 = 2Watt

Pc = 2E-3/1E-3 = 2Watt

8 0

2 years ago

Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the sun.

pogonyaev

Total mass of the Sun = 2x10^30kg

So 76% of that = (2x10^30kg)*(0.76) = 1.52x10^30kg ----> total amount of Hydrogen if only 12% of that is used for fusion, then (1.52x10^30kg)*(0.12) = 1.82x10^9kg

8 0

2 years ago

Complete the passage.

nataly862011 [7]

Answer:

Answer: Sound waves and some earthquake waves are longitudinal waves. Ocean, light and other earthquake waves are transverse waves.

Explanation:

There are 2 types of waves:

1. Longitudinal waves: These waves are defined as the waves in which the particles of the medium move in the direction of the wave. This requires a medium to travel. For Example: Sound Waves.

2. Transverse wave: These waves are defined as the waves in which the particles of the medium travel perpendicularly to the direction of the wave. This does not require a medium to travel. These can travel in vacuum also. For Example: Light waves.

Hence, Sound waves and some earthquake waves are longitudinal waves. Ocean, light and other earthquake waves are transverse waves

8 0

2 years ago