All categories

3 years ago

I need help with all the blanks

Physics

1 answer:

levacccp [35]3 years ago

6 0

There are only two out of the four squares that contain bb. So, 50% of the offspring will be brown. The other 50% of the offspring will be a hybrid because their combination is Bb. There is a 50% chance that all of the offspring could be born black.

You might be interested in

What is the independent variable

Ludmilka [50]

Answer:

independent variables are variables in mathematical modeling, statistical modeling and experimental science

6 0

2 years ago

Read 2 more answers

The electric field strength between the plates of a simple air capacitor is equal to the voltage across the plates divided by th

Dmitry [639]

Answer:

d = V/E

Explanation:

From the definition, we can say that the electric field strength between the plates of a parallel plate capacitor is

E = v/d

where

E = electric field strength

V = potential difference

d = distance between the plates

On rearranging the equation and making d subject of the formula, we have

d = V/E

From the question, we're given that

V = 112 V

E = 1.12 kV/cm converting to V/m, we have 110000 V/cm

d = 112 / 110000

d = 0.00102 m

d = 1.02*10^-3 m

5 0

2 years ago

what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?

zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

5 0

3 years ago

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99

sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

$\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]$

$\text{Average atomic mass}=20.169amu$

Hence, the average atomic mass of the given element is 20.169 amu.

4 0

3 years ago

How does Doppler ultrasound technology differ from ultrasound technology

patriot [66]

Answer:

. Doppler ultrasound is based on absorption of sound, and other

ultrasound technology is based on reflection.D.

Explanation:

6 0

2 years ago

Read 2 more answers