The most serious effect of radiation on humans is

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

The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}

tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

Explanation:

Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:

$K.E=\frac{1}{2}mv^2$

Kinetic energy of the 5 kg object.

Mass of object,m = 5 kg

Velocity of an object = v

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2$

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2$

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

$\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}$

Given that, v = 2v'

$\frac{K.E}{K.E'}=\frac{1}{1}$

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

3 0

3 years ago

A battery is used to power a flashlight. When the flashlight is in use, what type of energy is lost during energy transformation

diamong [38]

Answer:

The answer is chemical energy

4 0

3 years ago

Read 2 more answers

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

3 years ago

A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 672 nm. The slits have widths of 0.0

ololo11 [35]

Answer:

Explanation:

In case of diffraction , angular width of central maxima =2 λ/d

λ is wave length of light and d is slit width

In case of interference , angular width of each fringe

= λ /D

D is distance between two slits

No of interference fringe in central diffraction fringe

=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.

6 0

3 years ago